FIVE  YEARS 
QUESTIONS  AND  ANSWERS 

NATIONAL  ASSOCIATION  OF 
STATIONARY   ENGINEERS 


NATIONAL  ASSOCIATION/ 
STATIONARY  ENGINEERS 


Five  Years 

Questions  and 

Answers 

As  originally   published 

^^^^—^^^^—  in  — — — ^^^^~ 

The   National    Engineer 


VOLUMES     ONE     TO     FIVE     INCLUSIVE 


Marsh  &  Grant  Co 

Printers  and  Engravers 

Chicago 


Copyright  lyoz 

by 
GEORGE  D.   B.   VANTASSEL,   Sec'y 


P  R  E  F  A    C  E 


/TAHE  Questions  and  Answers  found  herein  were  first  published 
in  the  National  Engineer  in  a  competitive  educational  course, 
designed  to  stimulate  technical  work  and  theory  study  among  the 
members  of  the  National  Association  of  Stationary  Engineers. 

This  volume  becomes  the  fourth  bound  edition  of  the  same, 
which  represents  the  work  of  separate  committees  acting  through  a 
period  of  five  years.  These  committees  were  as  follows : 


F.  W.  ENLOY  .  . 
OTTO  LUHR  .  . 
CHAS.  W.  NAYLOR 


E.  J.  STODDARD 
E.  G.  JACQUES 

E.    P.    GlLROY 


CHAS.  H.  Fox  . 
ARTHUR  O.  HALL 
S.  J.  GRAIN  .  . 


M.  M.  CHILDS  . 
THOS.  P.  BURKE  . 
HY.  C.  HOFFMAN 


M.  M.  CHILDS     . 
GEO.  F.  HAVEN    . 


1896-1897 

.  Illinois  No.  29,  Chicago. 
.  Illinois  No.  38,  Chicago. 
.  Illinois  No.  28,  Chicago. 

1897-1898 

.  Michigan  No.  i,  Detroit. 
.  Michigan  No.  i,  Detroit. 
.  Michigan  No.  I,  Detroit. 

1898-1899 

.  Ohio  No.  36,  Cincinnati. 
.  Ohio  No.  15,  Cincinnati. 
.  Ohio  No.  2,  Cincinnati. 

1899-1900 

.  Rhode  Island  No.  I,  Providence. 
.  Rhode  Island  No.  2,  Pawtucket. 
.  Rhode  Island  No.  5,  Providence. 

1900- 1901 

.    Rhode  Island  No.   I,  Providence. 
.      .    New  York  No.  48,  Brooklyn. 


The  successful  associations  and  individuals  for  the  several 
years  were : 

For  1896-7:  Ohio  No.  36,  Cincinnati;  Iowa  No.  8,  Sioux 
City,  and  Illinois  No.  22,  Rockford. 

For  1897-8:  Ohio  No.  15,  Cincinnati;  Illinois  No.  22, 
Rockford,  and  Iowa  No.  8,  Sioux  City. 


For  1898-9:  Iowa  No.  8,  Sioux  City;  Louisiana  No.  i, 
New  Orleans,  and  Massachusetts  No.  17,  Lowell;  and  also  J.  S. 
Gillespie,  Pennsylvania  No.  12,  Philadelphia,  and  H.  H.  Carman, 
Ohio,  No.  28,  Akron. 

For  1899-1900:  New  York  No.  48,  Brooklyn;  Massachu- 
setts No.  17,  Lowell,  and  Ohio  No.  45,  Canton.  Also  for  elemen- 
tary work,  New  York  No.  48,  Brooklyn;  Ohio  No.  37,  Dayton, 
and  Michigan,  No.  24,  Detroit. 

For  1900-01:  Massachusetts  No.  17,  Lowell,  and  Massa- 
chusetts No.  1 6,  Waltham. 

Volumes  I,  II  and  III,  containing  the  problems  for  the  years 
1896-7,  1897-8  and  1898-9,  were  edited  by  Chas.  Desmond,  at 
that  time  editor  of  the  National  Engineer. 

This  volume  is  presented  by  the  undersigned  committee  in 
the  role  of  editors,  who  ask  your  kind  indulgence  for  their  effort, 
expressing  their  thanks  to  the  several  past  committees  for  assistance 
in  handling  the  proofs,  and  hopeful  that  the  book  may  attain  its 
object,  which  is  to  be  of  service  as  a  guide  and  reference  to  the 
practical  student-engineer. 

The  sequence  of  the  questions  and  answers  by  years  has  not 
been  respected  in  the  present  arrangement,  which  attempts  to  clas- 
sify the  several  problems  under  their  logical  headings.  The  result 
has  not,  however,  been  entirely  satisfactory  because  the  questions 
were  originally  presented  without  any  particular  thought  of  their 
ever  becoming  a  part  of  an  harmonious  whole.  A  large  demand 
for  the  book  will,  if  it  comes,  assure  the  committee  that  its  work 
has  been  fairly  well  done. 

CHAS.  W.   NAYLOR 

Illinois  No.  28,  Chicago. 

F.   ELMO  SIMPSON 

Illinois  No.  jj,  Chicago. 

GEORGE  NOWARD 

Illinois  No.  4.0,  Chicago. 


N  ATIONAL  ASSOCIATION    OF   STATIONARY    ENGINEERS 


QUESTIONS  w  ANSWERS 

1896-1901 


Boilers,  Furnaces,  Fuel,  Combustion, 
Chimneys,  Etc. 


Q.  1.  (1896-7.)  What  causes  worst  form  of  external  corrosion  of 
steam  boilers? 

Ans.  1.  External  corrosion  is  frequently  formed  by  exposure  of  the 
shell  to  the  cold  air.  Is  more  likely  to  occur  when  cold  than  when 
under  pressure  and  moisture  condensing  on  it  when  cold  aids  in  forma- 
tion of  rust.  It  is  likely  to  occur  along  line  of  brick  work  in  externally 
fired  boilers. 

Leaky  joints  are  another  source  of  corrosion,  either  from  riveted 
seams,  man  or  hand  holes,  or  from  imperfectly  fitted  attachments. 
Certain  coals  are  rich  in  sulphur,  the  products  of  whose  combustion 
contain  sulphurous  and  sulphuric  acid — this  has  a  bad  rusting  effect 
on  boilers.  [Wet  ashes  or  soot,  when  permitted  to  remain  in  contact 
with  the  boiler  plate,  cause  corrosion.] 


Q.  2.   (1896-7.)     Give  safe  strain  for  braces  in  steam  boilers. 
Ans.  2.     For   iron   6,000   Ibs.,   for  steel   7,000   Ibs.   per  square  inch 
cross-section. 


Q.  3.   (1896-7.)     Does  heat  or  pressure  cause  greatest  strain  in  boil- 
ers? 

Ans.  3.    Heat,  on  account  of  unequal  expansion. 


Q.  4.  (1896-7.)  How  to  construct  a  tubular  boiler  to  insure  free 
circulation? 

Ans.  4.  To  insure  free  circulation  in  horizontal  tubular  boilers  of 
the  ordinary  type  the  tubes  must  not  be  staggered  but  disposed  in 
straight  vertical  lines.  The  spacings  between  tubes  should  be  ample 


and  a  clear  space  of  4"  to  6"  is  proper  between  the  outer  row  of  tubes 
and  the  sides  of  the  boiler  shell,  together  with  a  space  from  14"  to  16" 
at  the  bottom.  The  upper  row  of  tubes  to  be  well  covered  when  the 
water  line  is  carried  at  a  point  a  little  above  2/3  of  the  shell  diameter. 
Some  designers  prefer  omitting  the  center  row  of  tubes  or  modifying 
the  space  to  allow  more  freedom  of  the  circulating  currents  at  that 
point. 


Q.  6.  (1896-7.)  Name  the  cheapest  and  most  commonly  used  scale 
preventer. 

Ans.  6.  Taking  the  country  as  a  whole,  sal-soda  is  the  cheapest 
and  most  commonly  used.  The  Eastern  section  is,  however,  using 
kerosene  oil  to  a  great  extent. 


Q.  7.  (1896-7.)  For  best  results  in  tubular  boiler  with  natural  draft, 
give  ratio  between  grate  and  heating  surface. 

Ans.  7.  This  would  depend  on  the  amount  of  draft  and  the  kind  of 
coal  used.  Slack  coal  requires  a  larger  grate  surface  than  lump  coal. 
Also  when  the  draft  is  strong  the  grate  area  does  not  need  to  be  so 
great  as  when  draft  is  poor.  All  chemists  appear  to  agree  that  the 
perfect  combustion  of  one  pound  of  coal  gives  exactly  the  same  number 
of  heat  units,  as  the  perfect  combustion  of  another  pound  of  the  same 
coal.  The  question  would  then  appear  to  resolve  itself  into  the  question, 
whether  we  can  get  nearer  the  perfect  combustion  with  strong  draft  and 
with  small  grate  surface,  or  with  a  moderate  draft  and  large  grate  sur- 
face. Ratios  vary  from  30:1  to  60:1.  Probably  45:1  is  the  best  ratio. 


Q.  11.  (1896-7.)     What  is  pitting?    How  caused? 

Ans.  11.  Pitting  is  conical  or  spherical  depressions  which  are  filled 
with  a  yellowish  brown  deposit,  consisting  mainly  of  peroxide  of  iron. 
Pitting  is  caused  by  the  action  of  oxygen  which  has  been  held  in  solu- 
tion by  the  water,  its  action  is  hastened  by  the  presence  of  carbonic  acid 
gas  which  is  liberated  when  the  temperature  of  the  water  is  increased. 
Boilers  that  are  kept  lukewarm,  and  in  which  the  circulation  is  poor, 
are  most  likely  to  be  affected  by  pitting. 


Q.  14.   (1896-7.)     Is  a  "mysterious  gas"  formed  in  boilers  at  times  of 
explosion? 

Ans.  14.  Its  existence  has  not  yet  been  proved,  although  some  be- 
lieve there  is  such  a  thing. 


Q.  17.  (1896-7.)     Give  relative  values  of  solid  boiler  plate,  double- 
riveted  and  single-riveted  joints;  who  established  them? 

Ans.  17.    Solid  plate  100   %,   double  riveted   70   %,   single  riveted 
ra    •  Wm'  Fairbairn-     [These  values  are  not  often  found  in 


Q.  20.  (1896-7.)     What  advantages  do   water   tube  boilers   possess 
over  other  types? 

b6  boilers  nave  the  followinS  advantages  over 


1.  Thin  heating  surface  in  boilers. 

2.  Joints  removed  from  the  fire. 

3.  Complete  combustion. 


4.  Large  draft  area. 

5.  Thorough  absorption  of  heat. 

6.  Efficient  circulation  of  water. 

7.  Quick  steaming. 

8.  Freedom  of  expansion. 

9.  Safety  from  explosions. 

10.  Ease  of  transportation. 

11.  Ease  with  which  they  can  be  repaired. 
[Some  of  these  statements  are  questionable.] 


Q.  21.  (1896-7.)  If  a  boiler  evaporates  3,000  Ibs.  of  water  per  hour, 
what  should  be  size  of  safety  valve?  (Pop  or  lever.) 

Ans.  21.  Rankine  says  the  constant  .006  X  the  water  evaporated 
per  hour  =  the  square  inch  area  of  the  required  safety  valve.  Then, 
.006  X  3,000  =  18"  area  and  4.8"  for  the  diameter  of  valve  needed. 

The  Commission  of  United  States  Supervisors  say  the  constant  shall 
be  .005.  Then,  .005  X  3,000  =  15"  area  or  4.375"  diameter  of  valve. 

Taking  it  that  one  square  foot  grate  surface  will  burn  12  Ibs.  of 
coal  per  hour  and  that  each  pound  of  coal  will  evaporate  8%  Ibs.  of 
water,  then  12X8%  =  102  Ibs.  of  water  will  be  evaporated  per  hour 
per  square  foot  grate  surface  and  3,000  -f-  102  =  29.4  sq.  ft.  grate  surface 
required.  With  the  ratio  of  one  square  inch  of  safety  valve  opening 
(pop  valve)  to  every  3  square  feet  of  grate  surface  we  have  29.4^ 
3  =  9.8  sq.  in.  required  in  valve;  or  a  diameter  of  3.53". 

With  a  ratio  of  one  square  inch  of  valve  opening  (for  lever  valve) 
to  each  2  sq.  ft.  of  grate  surface  we  have  29.4  -j-  2  =  14.7  sq.  in.  of  area 
in  valve  or  a  diameter  of  4.32". 

If  the  boiler  pressure  had  been  given  the  calculation  would  be  made 
on  the  basis  that  the  number  of  pounds  of  steam  that  will  flow  through 
an  orifice  ot  one  square  inch  area  in  one  second  may  be  found  by  divid- 
ing the  absolute  pressure  by  70.  The  following  formula  is  also  some- 
times used: 

%  water  evaporated  per  hr. 

Area  valve  — 

Steam  pressure  plus  10. 

Working  on  this  formula  with  assumed  pressures  as  below: 

1500 

50  Ibs.  B.  P.  = =  25     sq.  in.  area  =  5.65"  diameter. 

50+10 

1500 

100  Ibs.  B.  P.  = =  13.6  sq.  in.  area  =  5.17"  diameter. 

100+10 

1500 

150  Ibs.  B.  P.  = =  9.37  sq.  in  area  =  3.46"  diameter. 

150+10 


Q.  22.  (1896-7.)  Should  horizontal  externally  fired  boilers  be  set 
ievel  or  with  a  pitch?  If  with  a  pitch,  which  end  should  be  the  lower, 
and  why? 

Ans.  22.  First:  with  a  pitch.  Second:  back  end.  Third:  to  facili- 
tate the  draining  of  the  boiler  and  the  blowing  out  of  mud,  also  have 
more  heating1  surface  directly  exposed  to  fire,  and  as  a  safeguard  to  the 
water  column  by  keeping  more  water  at  the  back  than  is  indicated  in 
the  front  end  of  boiler. 


Q.  23.  (1896-7.)     Name  in  order  of  general  merit  the  four  most  com- 
monly used  boiler  metals. 

Ans.  23.    Steel,  wrought  iron,  copper,  cast  iron. 


Q  25  (1896-7.)  What  should  he  the  temperature  of  gases  in  an 
uptake  for  the  best  economy,  with  natural  draft  and  a  gage  pressure  of 
60  Ibs.? 

Ans  25  Theoretically  same  as  temperature  of  the  steam;  or  307° 
P.;  in  practice  from  50°  to  75°  or  sometimes  100°  greater  than  this, 
influenced  somewhat  by  draft. 


Q.  29.  (1896-7.)  How  many  tons  of  air  are  needed  to  burn  one  ton 
of  average  coal? 

\ns  29  Theoretically  about  12  tons,  but  in  practice  18  to  24  tons 
of  air.  (One  pound  of  coal  requires  12  to  24  Ibs.  air  to  burn  it.) 


Q.  32.     (1896-7.)     Can  smoke  once  formed  be  burned. 

Ans  32  Yes  and  no.  As  a  chemical  process  it  can  be  done,  but 
practically,  in  a  boiler  furnace,  we  think  not.  [Carbon  from  a  hydro- 
carbon is  the  sole  source  of  smoke;  heated  to  800°  or  upwards,  it  will 
combine  with  the  oxygen  of  the  air,  if  present  in  sufficient  quantity, 
and  smoke  will  be  prevented  in  the  furnace  or  consumed  if  such  condi- 
tions can  be  obtained  in  the  combustion  chamber  of  a  boiler  setting.] 


Q.  40.  (1896-7.)  What  should  be  the  pitch  of  rivets  to  give  the  best 
possible  percentage  of  strength,  in  a  staggered  double-riveted  lap  joint; 
thickness  of  plate,  5/16";  diameter  of  rivet,  11/16"  tensile  strength  of 
plate,  55,000  Ibs.;  shearing  strength  of  rivet,  38,000  Ibs.? 

Ans.  40.  In  a  riveted  joint  of  the  character  indicated  the  strain 
transmitted  by  the  load  carried  on  that  part  of  the  structure  repre- 
sented by  a  distance  equal  to  the  "pitch"  is  resisted  by  the  shearing 
strength  of  two  rivets. 

The  cross  section  of  a  single  rivet  11-16"  diameter  is  .3712  square 
inches;  hence  the  strength  of  the  joint,  as  far  as  the  riveting  is  con- 
cerned, is  found  by  doubling  the  area  and  multiplying  by  the  assumed 
shearing  strength  or  value  of  the  metal  per  square  inch  of  section,  viz. : 
2  X  .3712  =  .7424  X  38,000  =  28,211.2. 

The  best  possible  percentage  of  strength  in  any  riveted  joint  is 
realized  when  the  resistance  of  the  unimpaired  plate  section  between 
the  rivets  is  equal  to  the  strength  of  the  rivets.  To  obtain  this  result 
the  centers  of  the  rivets  must  be  proportioned  so  that  the  thickness 
of  the  metal,  taken  in  inches,  its  tensile  strength  per  square  inch  and 
the  length  of  the  effective  section  remaining  between  the  rivets,  when 
taken  together  as  factors,  produce  a  product  representing  a  sectional 
value  for  the  plate  equivalent  to  that  found  for  the  rivets. 

The  decimal  equivalent  of  5-16",  the  thickness  given,  is  .3125,  which, 
multiplied  by  55,000,  its  tensile  strength,  gives  17,187.5. 

The  value  of  the  third  factor  is  unknown;  therefore,  considering 
the  conditions  in  hand  in  the  form  of  an  equation,  we  have:  The 
resistance  of  the  rivets,  as  previously  found,  28,211.2,  giving  28,211.2 
=  17,187.5  X  D,  D  representing  the  unknown  factor,  from  which  we 
deduce  by  transposing:  D  =  28,211.2  -=-  17,187.5  =  1.641  inches.  Since 
D,  or  1.641,  represents  the  distance  between  rivet  holes,  the  actual  pitch 
is  one  diameter  of  the  rivet  more,  or  1.641  +  .687,  or  2.328  inches. 


Q.  41.  (1896-7.)  If  a  boiler  evaporates  5,000  Ibs.  water  per  hour 
from  and  at  212°,  how  many  HP.  will  it  be  rated? 

Ans.  41.  Horse-power  is  strictly  a  measure  of  work;  the  term  is 
conventionally  used  to  express  the  capacity  of  boilers. 

Since  the  actual  power  that  may  be  developed  by  any  given  volume 
of  steam  depends  upon  conditions  foreign  to  the  generator  producing 
it,  it  is  essential  that  we  assume  a  "base"  or  arbitrary  standard  in 
connection  with  the  expression  by  which  the  relative  capacities  of  dif- 
ferent boilers  may  be  intelligently  known,  without  reference  to  the 
duty  actually  performed  thereby.  An  evaporation  equivalent  to  30 
Ibs.  of  water  per  hour  taken  from  a  feed  temperature  of  100°  F.  into 
steam  at  70  Ibs.  gage  pressure,  as  fixed  by  a  board  of  experts  during 
the  Centennial  Exposition  in  1876,  is  generally  accepted  as  the  unit 
expressing  the  capacity  of  steam  boilers  in  horse-power. 

The  conversion  of  one  pound  of  water  into  steam,  under  the  con- 
ditions specified,  is  effected  by  the  absorption  of  1,110  thermal  units 
of  heat,  which  is  equivalent  to  the  evaporation  of  1110  -^  966  =  1.149 
Ibs.  at  a  temperature  of  212°  as  required.  Using  1.149  as  a  factor  and 
multiplying  the  same  by  30  gives  34.47,  which  represents  the  weight 
of  water  that  may  be  changed  to  steam,  from  and  at  212°,  by  the  same 
number  of  heat  units  required  to  effect  the  evaporation  of  30  Ibs.  of 
water  at  100°  into  steam  at  70  Ibs.  gage  pressure. 

The  quantity  34.37  is  usually  assumed  at  34.5  Ibs.,  hence  on  the 
basis  we  have  noted  the  direct  answer  to  the  question  may  be  found 
by  dividing  the  total  amount  evaporated  by  the  boiler  by  the  number 
which  represents  the  evaporation  required  for  one  horse-power,  giving: 
5000-^-34.5  =  144.92  HP.  [At  the  meeting  of  the  A.S.M.B.  a  boiler 
HP.  was  defined  as  34.5  Ibs.  water  evaporated  per  hour  into  steam,  from 
and  at  212°,  equivalent  to  the  transfer  of  34.5  X  965.7  =  33,317  B.T.U. 
per  hour.  See  part  of  Ans.  65  (1896-7).] 


Q.  42.  (1896-7.)  Where  should  the  feed  water  enter  the  boiler  and 
how  be  distributed,  in  the  best  practice,  and  for  the  best  results? 

Ans.  42.  A  prominent  authority  recommends  the  introduction  of  the 
feed  pipe  at  the  front  head,  just  above  the  upper  row  of  tubes,  and  ex- 
tending along  the  side  of  the  boiler  nearly  to  the  back  head.  It  then 
crosses  to  the  opposite  side  of  the  shell,  and,  turning  downward,  dis- 
charges between  the  shell  and  the  tubes.  [If  pipe  is  too  large  it  will 
fill  with  scale  until  force  of  water  is  sufficient  to  prevent  the  further 
formation  of  scale.] 

There  are  strong  arguments  in  favor  of  introducing  the  feed  water 
into  the  steam  space  in  the  form  of  a  spray,  this  preventing  in  a 
great  measure  the  formation  of  a  hard  scale  in  the  lower  parts  of  the 
boiler. 

In  any  case  the  comparatively  cool  feed  water  should  not  be  allowed 
to  come  into  contact  with  the  hot  part  of  the  shell  that  is  over  the  fire. 

It  is  calculated  that  if  a  plate  be  cooled  200°  a  longitudinal  strain 
of  8,000  Ibs.  to  10,000  Ibs.  per  square  inch  is  produced.  This,  in  addi- 
tion to  the  normal  strain  of  the  steam  pressure,  may  tax  the  seams 
beyond  their  elastic  limit. 

By  the  same  authority  it  is  given  out  that  girth  seams  develop 
leaks  and  cracks  in  99  cases  out  of  every  100  in  which  the  feed  dis- 
charges directly  upon  the  fire  sheets. 

Feeding  througn  the  blow-off  or  the  mud-drum  is  not  considered 
good  practice. 


Q.  43.   (1896-7.)     What  part  or  parts  of  a  horizontal  externally  fired 
boiler  are  subjected  to  greatest  strains  in  working? 


Ans.  43.  The  furnace  sheets  and  seams  over  or  near  the  fire  or 
bridge  wall  and  the  back  head  or  tube  sheet,  brought  about  through 
unequal  expansion  or  contraction.  Frequent  and  unnecessary  opening 
of  flue  doors  will  effect  front  head  and  flooding  the  boiler  with  cold 
feed  water  may  start  local  strains. 

Q.  46.  (1896-7.)  Sketch  to  a  scale  of  %  size,  the  head  of  a  48" 
tubular  boiler,  showing  the  thirty  4"  tubes,  and  a  4"  X  6"  handhole 
properly  placed. 

Ans.  46.    See  cut  on  following  page. 


Q.  47.  (1896-7.)  If  one  Ib.  of  coal  will  evaporate  10  Ibs.  water  from 
and  at  212°  F.,  how  many  pounds  will  it  evaporate 'from  80°  into  steam 
at  310°? 

Ans.  47.  The  total  heat  of  steam  is,  at  all  temperatures,  separable 
into  two  parts — latent  and  sensible  heat.  The  sensible  heat  is  that 
indicated  by  the  thermometer,  and  it  varies  as  its  pressure.  The  latent 
heat — absorbed  in  converting  water  into  steam — is  by  far  the  greater 
portion  of  the  total  heat.  Thus,  steam  at  a  temperature  of  212°  F.  has 
a  total  heat  of  1178.6  units,  so  that  in  evaporating  one  pound  of  water 
at  212°  into  steam  of  212°  we  have  added  1178.6 — 212°  =  966.6  units; 
to  evaporate  10  Ibs.  we  have  10  X  966.6  =  9666  units.  The  total  heat 
of  steam  at  310°  is  1208.5,  from  which  we  subtract  the  temperature  of 
the  water,  and  the  remainder  will  be  the  units  of  heat  required  to 
evaporate  one  pound  of  water  from  80°  into  steam  at  310°. 

1208.5—80  =  1128.5,  and  as  we  have  9666  units  to  apply,  then  9666  -f- 
1128.5  =  8.56  +  pounds  of  water. 


Q.  50.  (1896-7.)  A  ten-pound  piece  of  iron  is  left  in  the  gases  of  a 
chimney  until  thoroughly  heated  and  then  inserted  in  100  Ibs.  of  water 
at  50°  F.  The  resulting  temperature  of  the  water  and  submerged  iron 
is  55°.  Required:  The  temperature  of  the  chimney  gases. 

Ans.  50.  One  hundred  pounds  of  water  has  been  raised  5°  in  tem- 
perature; this  will  require  500  units  of  heat,  which  must  be  taken  from 
the  ten  Ibs.  of  iron  in  reducing  its  temperature  from  that  of  the  chimney 
gases  to  55°  F.  The  specific  heat  of  wrought  iron  is  .1138,  which  is  the 
amount  of  heat  it  will  be  necessary  to  put  into  1  Ib.  of  iron  to  raise  its 
temperature  1°,  or  the  amount  of  heat  that  lib.  of  iron  will  give  up  in 
falling  1°  in  temperature.  Therefore,  10  Ibs.  of  iron  v/ill  give  up  1.138 
heat  units  for  each  degree  drop  in  temperature,  then  500  -f-  1.138  =  439° 
fall  of  temperature,  and  439  +  55  =  494°  F.  as  the  temperature  of  the 
chimney  gases  in  the  supposed  case. 

Q.  63.  (1896-7.)  Name  five  things  or  conditions  upon  which  the 
efficiency  of  boiler  heating  surfaces  depend. 

Ans.  63.    1.  Difference  in  temperature  between  two  sides  of  the  plate 
L.    Thickness  of  plate. 

3.  Thermal  conductivity  of  plate. 

4.  Cleanliness  of  surfaces  inside  and  out. 

Hn     t  Ci.™ulation  of  water  in  boiler.     [Position  of  the  surface  in  rela- 
surface  ]       S°UrCe  °f  ^^     Angle  at  WMch  the  heat  rays  Strike  the 


setifn,t  Sh°uld  always  be  Put  °n  the  inside  of  the  fire 

causing  the  LtJhit6  ff^J6  •"  makes  a  place  for  sediment  to  gather! 
mg  the  patch  itself  to  be  injured  by  the  action  of  the  fire.    Interna 


SECTION  OF  BOILER  HEAD  SHOWING  ARRANGEMENT  OF 
TUBES  AND  BRACES 


pressure  will  hold  the  patch  in  place  and  not  have  a  tendency  to  blow 
It  off  the  boiler,  as  would  be  the  case  were  it  external.  The  calking 
area  is  reduced  by  putting  the  patch  on  the  inside  In  case  the  patch 
comes  at  the  edge  of  the  plate  it  will  be  necessary  to  let  it  come  on  the 
outside  of  the  adjacent  plate  on  account  of  the  difficulty  in  making  a 
tight  joint  if  it  was  on  the  inside  of  both  plates. 


Q  65  (1896-7.)  Which  is  the  more  accurate  way  of  measuring  the 
HP  of  a  boiler,  by  amount  of  water  it  will  evaporate  in  a  given  time,  or 
the  number  of  sq.  ft.  of  heating  surface  it  contains?  Give  reasons. 

Ans  65  In  measuring  the  horse-power  of  a  boiler  it  is  advisable  to 
assume  a  set  of  practically  attainable  results  in  average  good  practice, 
and  to  take  the  power  so  obtainable  as  a  measure  of  the  power  of  the 
boiler  in  commercial  and  engineering  transactions. 

The  unit  generally  assumed  has  been  the  weight  of  steam  demanded 
per  horse-power  per  hour  by  a  fairly  good  engine.  The  magnitude  has 
gradually  been  decreasing  since  the  early  times  of  the  steam  engine. 

in  the  time  of  Watt,  one  cubic  foot  of  water  per  hour  was  thought 
fair;  at  the  middle  of  the  present  century,  10  Ibs.  of  coal  was  a  good 
figure,  and  5  Ibs.,  commonly  equivalent  to  about  40  Ibs.  evaporation,  was 
allowed  for  the  engine. 

After  the  introduction  of  the  modern  form  of  the  engine  the  last 
figure  was  reduced  25%,  and  the  most  recent  developments  have  still 
lowered  this  consumption  of  fuel  and  steam. 

By  general  consent  the  unit  has  now  become  30  Ibs.  of  dry  steam  per 
hour  per  horse-power,  which  represents  the  performance  of  a  good  non- 
condensing  engine.  Large  engines,  with  condensers  and  compounded 
cylinders,  will  do  better. 

A  committee  of  the  A.S.M.E.  recommended  30  Ibs.  of  water  as  a  unit 
of  boiler  power,  and  this  is  now  generally  accepted.  They  advised  that 
the  commercial  rating  be  taken  as  an  evaporation  of  30  Ibs.  of  water 
per  hour  from  a  feed  temperature  of  100°  F.  into  steam  of  70  Ibs.  gage 
pressure,  which  may  be  considered  to  be  equal  to  34.5  units'  of  evapora- 
tion; that  is  34.5  Ibs.  of  water  from  a  feed  temperature  of  212°  into 
steam  at  same  temperature.  This  standard  is  equal  to  33,305  B.T.U. 
per  hour. 

A  boiler  may  have  a  large  heating  surface  and  not  be  as  efficient  as 
one  with  less.  Some  parts  of  heating  surfaces  are  more  effective  than 
others,  and  the  horse-power  per  square  foot  of  heating  surface  will 
vary  largely.  Therefore  we  think  the  proper  way  of  rating  a  boiler  is 
by  the  amount  of  water  it  will  evaporate  in  a  given  time.  Square  feet 
of  heating  surface  is  no  criterion  by  which  to  judge  different  styles  of 
boilers,  but  when  an  average  rate  of  evaporation  per  square  foot  for 
any  boiler  has  been  fixed  upon  by  experiment  this  becomes  a  more 
convenient  way  of  rating  the  power  of  other  boilers  of  the  same  style. 


Q.  69.     (1896-7.)     Is  scale,  or  deposit,  in  any  quantity,  a  good  thing 
in  a  steam  boiler?    Why? 

Ans.  69.  A  light  film  of  scale  is  believed  to  be  desirable  in  a  steam 
loiler.  It  is  well  known  that  a  free  contract  of  pure  water  is  destruc- 
tive to  boiler  plate.  The  same  may  be  said  of  the  acids  and  other 
:hemicals  commonly  introduced  with  the  feed  and  liberated  by  the 
LCtion  of  heat.  The  action  of  such  corrosive  influences  is  believed  to  be 
JSSST^K  ch^c^ed  by  the  interposition  of  a  thin  film  of  scale.  Some 
:  that  a  light  scale  prevents,  in  a  measure,  the  leakage  that  would 
occur  at  seams  of  a  poorly  built  boiler. 


Q.  71.  (1896-7.)  If  a  boiler  is  to  carry  a  working  pressure  of  105 
Ibs.  with  double  riveted  longitudinal  seams  of  70  per  cent  strength,  and 
using  a  factor  of  safety  of  5,  having  60,000  Ibs.  tensile  strength  of  plate, 
with  inside  diameter  of  shell  60",  what  should  be  the  thickness  of  the 
plates? 

Ans.  71.  To  find  the  thickness  of  boiler  plates  we  are  given  the 
following  formula: 

TX  SX  .70  =  PXRXF. 
When  T  =  thickness  of  plate  in  inches. 

S  =  tensile   strength   of   plate. 
.70  =  percentage  of  strength  of  joint. 
p  =  pressure  in  pounds  per  square  inch. 
R  =  radius  of  boiler  shell  in  inches. 
F  =  factor  of  safety. 
Transposing  above  formula  we  have: 

P  X  R  X  F        105  X  30  X  5       15750 

T  = = = =.375  =  %" 

S  X  .70  60000  X  .70          4200 

or  thickness  of  plate  required. 


Q.  84.  (1896-7.)  How  many  feet  of  rod  needed  to  put  three  guys 
on  a  smoke  stack,  if  fastened  55  ft.  above  the  ground  and  inclined  at  an 
angle  of  45°? 

Ans'.  84.  At  a  height  of  55  ft.  above  a  base  line  a  line  is  drawn  to 
an  angle  of  45°  from  the  vertical  line  and  a  line  drawn  from  this  point 
at  this  height  at  this  angle  would  pass  through  a  point  55  ft.  on  the 
base  line  from  its  intersection  with  the  vertical.  We  have  a  right  angle 
triangle  with  2  equal  sides.  To  find  the  hypothenuse.  This  equals  the 
square  root  of  the  sum  of  the  squares  of  the  two  sides  or  the  V552  +  552 
3=  V3025  +  3025  =  V6050  =77.78  +ft,  which  is  the  hypothenuse  or 
slant  side.  There  are  three  guys,  then  3  X  77.78  =  233.34  ft.  =  the 
length  of  the  guy  rod  needed,  making  no  allowance  for  eyes,  links,  or 
fasteners. 


Q.  89.  (1896-7.)  Is  the  use  of  the  steam  jet  as  an  auxiliary  to  fur- 
nace combustion  economical? 

Ans.  89.  A  committee  of  experts  in  St.  Louis  in  1891,  after  39  tests 
by  various  methods,  report  that  in  one  case  fuel  consumption  was  in- 
creased 12%  for  the  same  work.  Prof.  Landreth  then  said  "Steam  jets 
to  draw  air  in  or  inject  air  into  the  furnace  above  the  grate  and  also 
to  mix  the  air  and  combustible  gases  together  form  an  efficient  smoke 
preventer,  but  one  liable  to  be  wasteful  of  fuel."  From  above  and  also 
from  our  own  experience  we  incline  to  the  opinion  that,  generally 
speaking,  in  ordinary  practice,  a  jet  of  steam  above  the  grates  will  not 
increase  the  efficiency  of  a  boiler  as  a  steam  generator. 


Q.  90.  (1896-7.)  Given  a  lever  safety  valve:  Weight  of  ball  90  Ibs., 
distance  from  center  of  weight  to  fulcrum  40.41",  weight  of  lever  and 
valve  11%  Ibs.,  distance  from  fulcrum  to  center  of  gravity  of  lever  18", 
from  fulcrum  to  center  of  valve  4",  and  diameter  of  valve  3%";  at 
what  pressure  will  it  blow  off?  Explain. 

Ans.  90.  The  length  of  lever  being  40.41"  and  multiplying  this  by 
90  Ibs.,  or  the  weight  of  the  ball,  will  give  3639.9,  or  the  moment  of 
force  or  leverage  of  the  ball;  to  this  add  the  weight  of  the  valve-stem 
and  valve,  or  the  weight  of  the  valve-stem,  lever  and  valve,  by  its  ful- 
crum distance,  or  11%  X  18=211.5,  this  sum  +3639.9=3848.4,  gives  the  to- 
tal moment  tending  to  hold  the  valve  down,  against  which  we  have  a  cer- 


tain  pressure  trying  to  force  valve  up.  As  the  valve  is  3V2  diam t.  its 
area  is  9621  sq.  in ,  which  multiplied  by  4",  its  leverage,  gives  38.484 
as  its  leverage  moment.  This  divided  into  the  total  downward  force 
3848  4  gives  100  Ibs.  as  the  pressure  needed  to  balance  the  valve.  Prac- 
tically a  little  more  would  be  needed  to  make  the  valve  blow  off. 

Q  93  (1896-7.)  A  tubular  boiler  has  an  inside  diameter  of  60", 
and  is  to  carry  a  working  pressure  of  105  Ibs.  by  the  gage;  longitudinal 
seams  are  double-riveted,  70%  strength  of  joints,  plates  %"  thick,  factor 
of  safety  being  5,  what  should  be  the  tensile  strength  of  plates?  How 
found? 

Ans.  93.    Formula:        R  X  P  X  5  _ 

T  X  S 
Where  R  =  radius  in  inches. 

P  =  pressure  in  Ibs.  per  sq.  in. 
5  =  factor  of  safety. 
T  =  thickness  of  sheets  in  inches. 
S  =  strength   (efficiency)   of  joint. 
ts  =  tensile  strength. 

30  X.  105X5        15750 

.-. = =60000  ts. 

.375X70%        .26250 


NOTE:  These  lectures  to  be  used  in  explanation  of  Questions  1  and 
20  inclusive  of  1897-8  series. 

BOILER  AND  FURNACE. 

The  efficiency  of  the  furnace  and  boiler  is  always  directly  dependent 
upon  the  care,  skill  and  knowledge  of  the  engineer.  If  he  properly 
handles  the  coal  and  regulates:  the  supply  of  air,  the  coal  produces  the 
maximum  amount  of  heat  of  which  it  is  capable  and  the  boiler  absorbs 
the  greater  part  of  the  heat  produced  and  uses  it  in  the  production  of 
steam ;  if  he  does  not  properly  regulate  the  supply  of  coal  and  air,  either 
the  maximum  amount  of  heat  is  not  produced,  or,  if  produced,  the 
greater  part  is  carried  up  the  chimney  with  the  gases  and  is  radiated, 
or  both  these  sources  of  loss  occur  at  the  same  time. 

A  good  quality  of  coal  should  produce  about  14,000  heat  units  per 
pound,  when  properly  burned,  and  may  generate  as  little  as  3,500,  when 
improperly  burned  (Ans.  1  and  2).  Of  this  heat  the  boiler  should 
absorb  from  60  to  80  per  cent,  and  use  it  for  making  steam.  (Ans.  10.) 

Suppose  one  had  50  pounds  of  gases  passing  up  the  chimney  per 
pound  of  coal  burned,  and  these  gases  were  at  a  temperature  of  500° 
above  the  temperature  of  the  engine  room.  Then,  as  raising  one  pound 
of  the  gases  one  degree  takes  .24  of  a  heat  unit,  raising  it  500°  would 
take  .24  X  500,  or  120  heat  units.  Raising  50  Ibs.  to  that  temperature 
would  take  50  X  120,  or  6,000  heat  units.  Thus  he  is  sending  6,000 
of  his  14,000  available  heat  units,  or  43  %,  up  the  chimney,  and  he  could 
not  get  more  than  about  50  %  into  the  boiler,  as  about  7  %  would  be 
radiated  and  lost  in  other  minor  ways. 

Twenty-five  pounds  of  chimney  gases  per  pound  of  coal  is  sufficient 
(practical  trials  with  specially  skillful  firemen  generally  show 
MS)  and  the  average  temperature  of  the  chimney  gases  above  the  out- 
side air  in  the  137  tests  recorded  by  Mr.  Geo.  H.  Barrus  is  not  over 
JO  .  Taking  these  values,  only  about  17  %  of  the  heat  would  be  car- 
ried up  the  chimney  in  this  way. 

An  understanding  of  the  operation  of  the  furnace  and  boiler  requires 
that  we  should  know  the  following  facts: 

14 


(1)  The  heat  value  of  the  fuel. 

(2)  The  proportion  of  this  heat  used  in  making  steam. 

(a)  By  incomplete  combustion. 

(b)  By  hot  gases  passing  up  the  chimney. 

(c)  By  radiation  and  dropping  of  fuel  through  the  grate. 

(3)  The  proportion  of  this  heat  lost. 

The  second  and  third  added  together  should  equal  the  first.  Results 
may  be  checked  in  this  way: 

First: — In  the  answer  to  the  first  question  we  have  tried  to  collect 
data  that  shall  enable  us  to  ascertain  the  approximate  number  of  heat 
units  (quantity  of  heat)  that  should  be  produced  per  pound  of  fuel 
burned.  This  has  been  collected  from  papers  received  from  all  over 
the  country  in  reply  to  the  first  twelve  questions.  In  said  answer  the 
first  column  contains  the  name  of  the  coal,  the  second  the  number  of 
heat  units  that  will  be  produced  by  the  complete  combustion  of  one 
pound,  the  third  column  the  number  of  pounds  of  water,  from  and  at 
212°,  that  would  be  evaporated,  provided  60  %  were  used  for  this  pur- 
pose; the  fourth  column  the  same  as  the  third,  except  that  80  %  instead 
of  60  %  is;  here  assumed;  the  fifth  column  gives  the  pounds  of  water, 
from  and  at  212°,  that  have  been  evaporated  by  one  pound  of  some  of 
the  coals  mentioned,  in  actual  practical  tests;  the  sixth  column  gives 
the  per  cent  efficiency,  corresponding  to  the  actual  evaporations  named 
in  column  five. 

Second: — The  amount  of  water  evaporated  during  a  given  time  may 
be  estimated  as  indicated  in  question  No.  21  or  in  various  other  ways. 
This  amount  in  pounds  divided  by  the  number  of  pounds  of  coal  burned 
during  the  same  time  will  give  the  evaporation  per  pound  of  coal. 
This  multiplied  by  the  number  of  heat  units  required  to  evaporate  one 
pound  will  give  the  amount  of  heat  utilized  in  making  steam.  Or  the 
evaporation  may  be  reduced  to  an  equivalent  evaporation  from  and  at 
212°  by  multiplying  it  by  the  proper  number  (factor  of  evaporation) 
contained  in  the  following  table  abridged  from  "Kinealy  on  Engines 
and  Boilers."  This  product  multiplied  by  966  will  give  the  quantity 
of  heat  used  in  forming  steam.  Dividing  the  heat  value  of  the  coal 
by  this  last  product  will  give  the  efficiency  of  the  boiler  and  furnace. 

Temperature      Pressure  of  Steam  by  Gauge  in  Ibs.  per  Square  Inch, 
of  Feed  Water.  Factors  of  Evaporation. 

O.P.    40P.    50P.    60P.    70P.    SOP.    90P.  100P.  HOP.  120P. 

40  1.179  1.203  1.206  1.209  1.212  1.214  1.217  1.219  1.221  1.222 

60  1.158  1.182  1.185  1.188  1.191  1.193  1.196  1.198  1.200  1.201 

70  1.148  1.172  1.175  1.178  1.181  1.183  1.186  1.188  1.190  1.191 

90  1.127  1.151  1.154  1.157  1.160  1.162  1.165  1.167  1.169  1.170 

110  1.106  1.130  1.133  1.136  1.139  1.141  1.144  1.146  1.148  1.149 

130  1.085  1.109  1.112  1.115  1.118  1.120  1.123  1.125  1.127  1.128 

150  1.065  1.089  1.092  1.095  1.098  1.100  1.103  1.105  1.107  1.108 

170  1.044  1.068  1.071  1.074  1.077  1.079  1.082  1.084  1.086  1.087 

190  1.023  1.047  1.050  1.053  1.056  1.058  1.061  1.063  1.065  1.066 


TEST  OF  BOILER  AND  FURNACE. 

A  complete  test  of  a  boiler  and  furnace  would  involve  ascertaining 
the  following  facts:  — 

First.     The  quantity  of  heat  generated  by  the  combustion  of  the  coal. 
Second.    The  heat  used   in   making  steam. 
Third.     The  proportion  of  the  heat  lost  by — 

(a)  Incomplete  combustion. 

(b)  Hot  gases  passing  up  the  chimney. 

(c)  Radiation  and  dropping  of  fuel  through  the  grate. 

The  second  divided  by  the  first  is  the  efficioncy  of  the  furnace  and 
boiler. 

15 


flso  eaPsy  to  fstimatewhetheror  not  the  loss  from  incomplete  combus- 

U° "/thTfuenfnot  evenly  distributed  on  the  grate  one  may  have  too 
much  air  supplied  to  a  part  of  the  furnace  and  not  enough  to  the  re- 
mainder, so  that  excessive  quantities  of  free  oxygen  (0.)  and  of  carbon 
monoxide  (CO)  in  the  chimney  gases  would  indicate  both  an  excessive 
amount  of  air  and  also  incomplete  combustion. 

If  the  remaining  sources  of  loss  are  very  considerable  it  is  owing 
to  inexcusable  mismanagement  or  imperfect  apparatus.  One  author 
remarks  that  the  remedy  is  to  get  a  new  fireman. 

The  following  gives  a  tabulated  result  of  an  1897  practical  test: 
Coal  Used  Pocahontas  (run  of  mine)  ;  Heat  Value,  14,289. 

Per  cent  of 
Heat  units.          total  heat. 

Useful  evaporation   11374'C!40t 

Loss  by  chimney  gases 1900.437  16.6 

Unconsumed  carbonic  oxide 271.491  1.9 

Loss  by  radiation,  unconsumed  hydrocarbons, 
evaporation  of  moisture  in  coal  (2.3%)  and 
unaccounted  for •  743.028  5.2 

Total..  14289  100 


The  Quantity  of  Heat  Generated. 

Mr.  R.  S.  Hale,  in  his  paper  on  "Fuel  Gas  Analysis  in  Boiler  Tests," 
read  before  the  American  Society  of  Mechanical  Engineers  in  1897, 
remarks  that  it  is  not  necessary  to  have  an  analysis  of  the  coal,  inas- 
much as  the  data  of  the  text  books  can  be  relied  upon  within  very 
narrow  limits.  In  answer  to  Q.  1,  your  committee  has  endeavored  to 
supply  reliable  data;  of  course  it  is  open  to  correction. 

To  find  the  quantity  of  heat  that  should  have  been  generated  in  the 
furnace,  one  would  look  in  the  table,  Ans.  1,  for  the  kind  of  coal  he  was 
using,  and  opposite  that  he  would  find  the  number  of  heat  units  gen- 
erated per  pound  of  coal  burnt.  This  multiplied  by  the  number  of 
pounds  of  coal  burned  in  a  given  time  would  be  the  total  heat  that 
should  have  been  produced  during  that  time.  The  coal  is  usually 
weighed  in  the  barrow  upon  platform  scales. 


The  Proportion  of  Heat  Used  in  Making  Steam. 

In  the  report  of  the  Committee  on  Code,  at  this  year's  meeting  of 
the  A.S.M.E.,  occur  these  words:  — 

"The  elaborate  directions  and  multiplicity  of  details  provided  for  in 
the  foregoing  code  should'  not  divert  the  mind  from  the  fact  that  the 
principal  elements  to  be  ascertained  in  a  boiler  test  are  the  weight  of 
water  evaporated  and  the  weight  of  the  fuel  required  to  produce  such 
evaporation." 

The  water  fed  to  the  boiler  is  best  weighed  in  a  separate  tank, 
wnich  is  emptied  into  a  tank  from  which  the  boiler  draws  its  supply. 

The  amount  may  be  approximately  estimated  as  indicated  in  Ans. 

it  would  perhaps  pay  to  actually  measure  the  water  necessary  to 

draw  from  the  boiler  to  make  the  level  drop  from  one  mark  to  another 

the  water  glass  when  the  boiler  was  not  in  use.    It  might  also  be 
measured  by  a  water-meter. 
th^^Sf  g?fc  ,t.he ,water  evaporated,  this  amount  would  be  divided  by 

J  number  indicating  the  weight  of  coal  used  and  the  result  reduced 
equivalent  evaporation  from  and  at  212°  by  multiplying  by  the 

16 


factor  of  evaporation  taken  from  one  of  the  numerous  tables  published. 
The  above  last  result  would  then  be  multiplied  by  966,  and  the 
product  divided  by  the  heat  value  of  the  coal,  to  get  the  efficiency 
of  the  furnace  and  boiler,  which  should  be  from  60  %  to  80  %. 


ANALYSIS  OF  FLUE  GASES. 

As  a  check  upon  the  above  results,  also  a  reliable  indication  of  the 
performance  of  the  furnace,  we  wish  to  know  two  things:  — 

First.  How  many  pounds  of  gas  are  going  up  the  chimney  per 
pound  of  coal  burned,  together  with  its  temperature? 

Second.  Whether  or  not  a  considerable  amount  of  combustible  gag 
is  going  up  the  chimney? 

The  first  may  be  determined  approximately  by  the  rules  given  in 
Ans.  15. 

A  rough  but  practical  estimate  of  the  second  may  easily  be  made 
when  we  recollect  that  air  is  about  20  %  oxygen  by  volume  and  80  % 
nitrogen;  that  the  nitrogen  goes  through  the  furnace  unchanged;  that 
as  much  of  the  oxygen  as  is  properly  used  is  changed  to  an  equal  vol- 
ume of  CO*  (carbonic  acid).  Therefore  if  all  the  oxygen  of  the  air 
is  either  unchanged  or  is  changed  to  carbonic  acid,  the  volume  of  the 
carbonic  acid  and  oxygen  taken  together  must  still  be  20  %,  so  that  if 
we  measure  the  volume  of  carbonic  acid  and  oxygen,  find  what  per 


cent  they  are  of  the  whole,  taken  together,  subtract  this  result  from 
20,  we  shall  have  as  a  remainder  a  per  cent  which  is  roughly  propor- 
tional to  the  combustible  gases  in  the  chimney.  If  this  remainder  is 
greater  than  2  (two)  it  would  generally  indicate  that  insufficient  air  is 
being  supplied  to  the  furnace.  In  all  ordinary  cases  the  second  is 
neglibly  small. 


For  the  volumetric  analysis  of  the  chimney  gases  the  Orsat  appa- 
ratus is  generally  recommended  as  accurate,  convenient  and  reliable. 
Its  cost  is  somewhere  from  $16  to  $30.  The  form  using  pinch-cocks 
instead  of  glass-cocks  is  recommended  by  Prof.  Gill  as  simpler, 
cheaper  and  more  practical.  Of  this  apparatus  the  Committee  on  Code 
of  the  A.S.M.E.  says:  — 

"For  the  past  year  the  writer  has  made  extensive  use  of  the  Orsat 
apparatus  in  his  boiler  testing,  and  has  found  the  work  not  only  inter- 
esting but  exceedingly  instructive  and  valuable.  Its  chief  value  lies 
in  the  guide  which  it  affords  in  determining  what  kind  of  firing  is 
most  advantageous  where  the  fuel  is  bituminous;  coal.  That  the  instru- 
ment is  reliable  and  useful  for  the  purpose  noted  is  quickly  ascertained 
and  without  any  very  extended  practice.  When  the  thickening  up  of 
the  fire  is  invariably  attended  with  an  increase  in  the  percentage  of 
carbonic  oxide  and  a  reduction  in  the  percentage  of  oxygen,  as  the 
writer  has  found,  he  feels  at  once  assured  that  the  instrument  is'  not 
a  plaything  or  something  that  is  influenced  in  unexplained  ways  by 
whim  or  caprice,  but  rather  that  it  is  an  important  adjunct  to  the 
engineer's  outfit." 


HOME-MADE  ANALYZING  TUBE. 

The  device  described  below,  which  has  been  constructed  and  used 
by  your  committee,  is  upon  the  same  principle  as  the  Orsat  apparatus, 
but  is,  of  course,  not  so  elaborately  constructed  and  graduated.  This 
apparatus  is  illustrated  in  Figs.  1  and  2.  Anyone  can  make  it,  and  the 
apparatus  itself  ought  to  cost  about  50  cents. 

Fig.  1. — A  is  a  stick  held  in  a  vertical  position  by  a  vise.  B  C  is 
a  small  cup.  D  is  a  graduated  measuring  glass;  it  is  immaterial  into 
what  units  the  glass  is  graduated,  so  that  the  unit  is  small.  E  is  a 
glass;  tube  %"  or  more  in  diameter  and  about  15"  long,  drawn  down 
in  a  gas  flame  at  the  ends,  so  that  a  small  rubber  (black  preferred) 
tube,  as  F,  may  be  slipped  on  and  make  air-tight  joints.  G  is  a  pinch- 
cock  by  which  the  tube  F  may  be  closed  air-tight.  H  is  a  funnel;  in 
this  case  it  is  the  part  of  the  tube  E  that  was  separated  from  the  rest 
when  the  ends  were  drawn  down.  I  is  a  rubber  band,  which  is  placed 
around  the  stick  A  and  funnel  H  to  hold  the  apparatus  in  a  vertical 
position.  J  are  rubber  bands  upon  the  tube  E.  These  may  be  moved 
to  different  positions  on  the  tube,  to  serve  as  marks.  If  a  graduated 
tube  is  used  the  glass  D  may  be  dispensed  with. 

METHOD   OF   USING. 

The  pinch-cock  G  is  opened  and  the  tube  E  filled  with  water,  which 
is  then  allowed  to  run  out  slowly  into  the  graduate  and  measurer.  The 
measurement  of  this  water  determines  the  volumetric  contents  of  the 
tube  once  for  all.  Suppose,  for  convenience,  it  is  100  cubic  centimeters 
(though  it  may  be  anything;  the  larger,  the  more  accurate  will  be  the 
measurements) . 

It  is  generally  thought  best  to  lead  a  W  or  %"  lead  pipe  from  the 
"breaching"  or  a  part  of  the  flue  near  the  up-take  to  a  convenient  place 
for  filling  the  tube.  We  have  used  a  small  rubber  hose.  It  is  said  that 
if  rubber  gets  too  hot  it  will  generate  gases  that  will  effect  the  results. 

To  get  the  gases  into  the  tube  (see  Fig.  3)  one  simply  connects  the 
;  C,  in  line  with  the  pipe  A,  to  the  breeching  X,  and  also  to  con- 

t  said  tube  to  the  inlet  port  of  a  hand  air-pump  B,  and  draws  a 
charges  of  the  pump  through  the  tube.  This  will  clear  out  the 
Lir  that  was  in  the  tube  and  replace  it  with  flue  gases.  Perhaps  it 
might  be  better  to  take  a  larger  number  of  strokes  of  the  pump. 

A  sort  of  ejector,  like  that  shown  in  Fig.  4,  may  be  used  instead  of 
a  pump. 


By  the  aid  of  a  small  jet  of  steam  a  steady  stream  of  flue 
may  be  drawn  through  the  tube. 

A  barrel,  keg,  or  large  can,  filled  with  water,  may  be  used  for  this 
purpose,  as  indicated  in  Fig.  5,  in  which  C  is  the  tube  connected  with 
the  top  of  the  barrel,  and  with  the  pipe  A  (Figs.  3  and  5).  Letting  the 
water  run  out  of  the  bottom  of  the  barrel  draws  the  flue  gases  through 
the  tube  at  the  top.  Care  should  be  taken  not  to  let  the  water  get 
low  enough  so  that  air  will  be  drawn  back  by  the  draft. 

The  tube  E  is  connected,  by  means  of  rubber  tubes  upon  its  ends, 
with  a  pipe  or  tube  leading  to  the  chimney,  and  the  flue  gases  are 
drawn  through  by  means  of  an  air  pump  or  other  convenient  device 
until  all  the  air  is  removed  and  the  tube  is  full  of  chimney  gases. 

Having  got  the  gases  into  the  tube,  the  pinch-cock  G  is  closed,  the 
other  end  of  the  tube  B  is  closed,  say,  by  the  finger  (Fig.  2).  The 
cup  C  is  about  two-thirds  full  of  caustic  potash  solution.  The  finger 


and  end  of  the  tube  E  are  now  put  into  the  cup  and  the  finger  removed 
from  the  end  of  the  tube  while  both  are  under  the  surface  of  the  fluid. 
The  apparatus  is  now  in  the  position  shown  in  Fig.  1,  though  the  fun- 
nel H  need  not  be  on.  The  pinch-cock  G  is  now  opened  for  an  instant 
until  some  of  the  fluid  in  the  cup  has  run  into  the  tube,  and  a  rubber 
band,  J,  moved  down  to  mark  the  surface  of  the  fluid  in  the  tube.  The 
finger  is  again  placed  over  the  end  of  the  tube  (under  the  surface)  and 
the  tube  turned  up  so  that  the  fluid  will  run  along  the  walls  of  the 
tube  and  expose  a  good  deal  of  surface  to  the  contained  gas,  as  shown 
in  Fig.  2.  Upon  placing  the  tube  E  back  into  the  cup  in  the  position 
of  Fig.  1  it  will  be  noticed  that  the  fluid  is  drawn  up  into  the  tube. 

Repeat  the  above  operation  three  or  four  times  at  intervals  of  about 
a  minute,  or  until  the  fluid  no  longer  rises.  Push  another  of  the  bands 
J  down  to  mark  the  surface  of  the  fluid  in  the  tube. 


The  distance  between  the  two  rubber  bands  measures  the  volume 
of  carbonic  acid  gas  (CO2)  that  was  in  the  tube;  this  volume  divided 
by  the  total  volume  of  the  tube  above  the  first  band  is  the  per  cent 
of  carbonic  acid  gas  in  the  flue  gases. 

Now  leaving  everything  in  position,  put  the  funnel  H  in  the  end 
of  the  rubber  tube  F,  as  shown.  Then  fill  the  glass  graduate  or  any 
convenient  vessel  about  two-thirds  full  of  water  and  put  into  it  all  the 
snow-like  pyrogallic  acid  that  it  will  absorb.  Fill  the  funnel  with  this 
pyrogallic  acid  solution  and  carefully  open  the  pinch-cock  G  a  very 
little,  allowing  a  little  of  the  contents  of  the  funnel  to  run  down.  Be 
careful  and  not  overdo  this;  enough  to  fill  the  tube  *4"  to  W  is  suf- 
ficient. On  repeating  the  operation  indicated  by  and  described  in  refer- 
ence to  Fig.  2  it  will  be  noticed  that  the  fluid  in  the  tube  begins  to 
turn  a  dark  orange  color  and  the  level  to  rise.  This  indicates  the 
absorption  of  oxygen. 

When  the  level  no  longer  rises,  move  another  rubber  band  down  to 
mark  its  position.  The  distance  between  the  second  and  third  rubber 
bands  measures  the  free  oxygen  that  was  in  the  tube  and  has  been 
absorbed;  this  volume  divided  by  the  volume  of  the  tube  above  the  first 
band  is  the  per  cent  of  free  oxygen  in  the  flue  gases. 

As  long  as  the  rubber  bands,  Jl,  J2,  J3,  are  not  moved,  one  may 
measure  the  volume  of  the  tube  between  them  as  often  as  he  pleases 
by  drawing  water  up  into  the  tube  and  allowing  it  to  flow  out  into 
the  graduate  until  it  falls  from  one  band  to  the  other.  If  one  wants 
very  great  accuracy  he  should  calculate  the  effect  of  the  column  of 
fluid  drawn  up  into  the  tube,  and  should  be  careful  to  let  the  tube 
drain  long  enough  before  measuring,  and  also  consider  the  pressures 
of  the  water  vapor.  The  temperature  of  the  tube  and  contents  must 
not  vary  during  the  operation. 

The  finger  must  never  be  removed  from  the  end  of  the  tube  E, 
except  when  under  the  surface  of  the  fluid  in  the  cup  C. 

Unless  one  has  particularly  thin  skin,  or  unless  the  operation  is 
unnecessarily  protracted,  the  caustic  will  do  no  harm.  Running  water, 
olive  oil  and  muriatic  acid  will  neutralize  or  remove  the  potash. 

The  writer  has  protected  his  finger  by  placing  over  it  a  thin  rubber 
bag  taken  from  the  toy  technically  called  a  "cry-baby,"  sold  at  most 
of  the  shops  for  a  nickel.  Our  brethren  in  New  York  and  Chicago  may 
be  able  to  obtain  something  better  for  the  purpose — something  specially 
made  to  resist  the  action  of  corrosive  fluids.  [Rubber  finger-stalls  can 
be  purchased  from  dealers  in  rubber  goods. — Ed.] 

COST  OF  APPARATUS. 

Glass  tube   $0.10 

1  ft.  black  rubber  hose 0.10 

Glass  graduate  0.25 

Pinch-cock    0.10 

1  Ib.  caustic  potash 0.10 

1  can  pyrogallic  acid...  ..  0.35 


Total   $1.00 


ESTIMATING  THE  WEIGHT  OF  THE  CHIMNEY  GASES. 

For  this  purpose  it  would  only  be  necessary  to  ascertain  the  per  cent 
by  volume  of  the  carbon  dioxide.  Ordinarily  the  weight  of  the  gases  per 
pound  of  coal  would  run  about  as  follows  for  the  corresponding  per 


20 


Weight  of 

Per  cent  Flue  Gases 

carbon  dioxide.  per  Ib.  coal. 

3.  61 

4.  50 

5.  40 

6.  33.3 

7.  28.6 

8.  25 

9.  22.2 

10.  20 

11.  18.2 

12.  16.7 

13.  15.4 

14.  14.3 

15.  13.3 

16.  12.5 

Under  the  conditions  at  each  end  of  the  column  there  would  prob- 
ably be  considerable  inflammable  gases  in  the  chimney  gases.  (See 
Ans.  16  and  remarks  under  the  heading  "Analysis  of  flue  gases.") 
That  would  have  to  be  estimated  as  previously  described.  For  the  pur- 
pose of  finding  the  volume  of  CO2  only,  the  tube  would  not  haye  to  be 
manipulated,  nor  would  it  be  necessary  to  put  the  hand  nor  any  part 
of  it  into  the  alkali.  A  little  of  the  potash  solution  would  be  put  in 
the  cup,  the  funnel  would  be  filled  with  it,  and  a  little  let  down  occa- 
sionally by  opening  the  pinch-cock  so  as  to  keep  the  walls  of  the  tube 
covered  with  the  solution.  The  correction  for  the  weight  of  the  col- 
umn in  the  analyzing  tube  is  made  as  above  described  in  referring  to 
the  oil  column  except  that  the  constant  would  be  .0063. 

AN   ILLUSTRATIVE   TEST. 

Duration  of  test,  10  hours. 

Kind  of  coal  used,  Pocahontas.    Heat  value,  14,290. 

Weight  of  coal  consumed,  6,507  Ibs. 

Water  fed  to  boiler,  64,877  Ibs. 

Water  per  pound  of  coal,  64,877  -^  6,507  =  9.97  Ibs. 


Equivalent  evaporation  from  and  at  212°  9.97  X  1.19  =  11.80  Ibs. 
Heat  used  in  making  steam  per  pound  of  coal  used,  11.8  X  966  = 
11,398  B.  T.  U. 

Efficiency  of  boiler  and  furnace,  11,398  4- 14,290  =  .80. 
Gas  analysis  per  cent  by  volume — carbonic  acid  15.1,  oxygen  4. 
Leaving  20 — 19.1  =.9  %,  indicating  a  small  amount  of  combustible 
gas  in  chimney. 

Weight  of  chimney  gases  per  pound  of  coal,  11  X  (15.1  +  4)  -r- 15.1 
=  13.92. 

Temperature  of  flue  gases  above  outside  air,  529°. 

Sensible  heat  in  flue  gases,  529  X  13.92  X  .24  =  1,767  heat  units. 

Per  cent  of  heat  in  flue  gases,  1,767  -f-  14,290  =  12.3. 

Heat  Units.        Per  Cent. 

Steam  making 11,398  80 

Hot  gases    1,767  12.3 

Balance:  radiation,  unconsumed 
gases,  dropping  coal  through 
grate,  etc 1,125  7.7 


Total    14,290 


100 


FIG.  6. 


FIG.  4. 


FIG.  5- 


FIG.  7. 


For  practice  in  analyzing  gases  one  may  find  the  proportion  of 
carbonic  acid  and  oxygen  in  his  breath.  It  is  better  to  hold  the  breath 
as  long  as  convenient  before  blowing  it  through  a  tube,  and  to  blow 
through  the  tube  enough  times  to  be  sure  that  all  the  air  originally  in 
the  tube  is  expelled. 


TEMPERATURE  OF  CHIMNEY  GASES. 

We  have  felt  that  the  methods  of  ascertaining  the  temperature  of 
chimney  gases  were  not  very  available  and  practical  for  our  purposes. 
Below  will  be  found  described  a  device  for  that  purpose,  which  we  be- 
lieve accurate,  simple  and  convenient. 

The  drawings  are  the  work  of  one  of  the  committee,  and  are  in- 
tended simply  to  be  illustrative,  and  no  great  accuracy  nor  proportion 
in  sizes  have  been  striven  for. 

We  have  experimented,  successfully,  we  believe,  with  the  following 
described  apparatus: 

In  Fig.  6,  A  represents  the  front  wall  of  the  breeching;  B  is  a  glass 
tube,  say  1/4"  outside  diameter,  and  as  long  as  can  conveniently  be  se- 
cured, drawn  together  at  one  end,  b,  in  a  gas  flame;  the  other  end, 
which  is  open,  or  slightly  contracted,  stuck  through  a  cork  or  plug, 
C.  The  cork,  C,  is  then  forced  into  the  hole  in  the  breeching,  so  that 
the  glass  tube  extends  into  the  hot  flue  gases,  as  shown,  and  left  there 
until  it  has  the  same  temperature  as  the  gases. 

It  Is  then  taken  out  and  placed  with  its  open  end  under  the  surface 
of  some  high  boiling  point  (flashing  point)  oil  (we  use  engine  oil) 
as  shown  in  Fig.  7,  and  left  there  to  cool.  When  it  has  cooled  down 
to  the  temperature  of  the  room,  the  oil  will  have  risen  up  into  the 
tube  a  certain  distance,  which  is  proportional  to  the  amount  the  col- 
umn of  air  in  the  tube  has  contracted  in  cooling,  and  therefore  pro- 
portional to  the  change  in  the  temperature  of  the  tube.  The  column 
of  air,  b,  e,  Fig.  7,  is  now  measured.  Inasmuch  as  the  volume  of  air 
is  proportional  to  its  absolute  temperature  and  the  volume  of  air  in 
the  tube,  b,  e,  is  at  the  temperature  of  the  room  and  equal  to  the  length 
of  the  tube  at  the  temperature  of  the  flue  gases,  therefore,  b,  e,  is  to 
the  length  of  the  tube  as  the  temperature  (absolute)  of  the  room  is  to 
the  absolute  temperature  of  the  flue  gases. 

The  tube  we  used  was  15.5"  long.  The  temperature  of  the  room  was 
90°;  that  is  460  +  90  =  550°  absolute.  The  oil  was  drawn  up  7",  leav- 
ing 15.5  —  7  =  8.5"  as  the  length  of  the  column,  b,  e,  above  the  oil. 
Then  8.5  :  15.5  : :  550  :  the  absolute  temperature  of  the  flue  gases. 
Therefore  the  absolute  temperature  of  the  flue  gases  was  550  X  15.5  -f- 
8.5  =  1003  or  in  ordinary  measurement  1003  —  460  =  543°.  This  would 
be  543  —  90  =  453°  above  the  outside  air. 

The  weight  of  the  column  of  oil  has  a  slight  stretching  out  effect 
upon  the  air  above  it,  because  of  its  weight.  To  find  how  much  should 
be  added  to  the  height  of  the  column  for  this  effect,  one  multiplies  the 
height  of  the  column  of  oil  by  the  length  of  the  tube  above  the  oil  and 
this  product  by  .004;  the  result  is  the  fraction  of  an  inch  that  should 
be  added  to  the  height  of  the  oil  column. 

In  the  above  example  the  correction  would  be  as  follows:  7  X  8.5  X 
.004  =  .238.  Therefore  to  the  measured  height  of  the  oil  column 
should  be  added  about  .24  of  an  inch. 

One  might  use  a  metal  tube,  for  instance  a  piece  of  gas  pipe,  with 
one  end  closed  air  tight  and  the  other  nearly  closed,  instead  of  the  glass 
tube,  if  he  prefers;  but  in  this  case  he  must  have  a  measuring  glass 
(or  a  pair  of  scales)  to  find  the  volume  of  the  air  in  the  tube  origin- 
ally,, and  the  volume  after  contraction.  For  example:  Suppose  we 
had  a  tube  of  2"  inside  diameter,  and  5"  long.  We  will  fill  it  with 
water,  then  pour  the  water  out  and  measure  it,  and  find  that  it  is  15.5 
cu.  ins.  Then  we  put  the  tube  into  the  breeching  and  let  it  get  as  hot 

23 


In  Fig.  8  we  have  drawn  our  idea  of  this  last  form:  A  is  a  hook  to 
hang  it  by  in  the  flue  or  chimney.  Of  course  the  tube  must  be  perfect- 
ly dry  when  placed  in  the  flue  and  no  steam  in  it.  We  do  not  think  the 
evaporation  of  the  water  will  effect  the  result. 

When  metal  is  used  it  would  be  well  to  have  pretty  thick  walls  so 
that  the  interior  would  not  cool  off  before  one  could  get  it  into  the 
water.  We  would  remark  that  the  method  of  placing  the  tube  in  the 
chimney  is  a  matter  of  personal  choice  and  judgment. 

COMPRESSED  AEB. 

The  usual  formula  in  the  book  is 

PVi-*=C 

in  which  P  is  the  pressure,  V  the  volume,  and  C  a  constant  which  has 
to  be  calculated  for  each  compressor. 

This  formula  is  somewhat  difficult  to  use  because  it  has  a  decimal 
fraction  for  the  exponent  of  V.  Moreover,  it  assumes  that  no  heat 
is  lost  by  the  air  which,  owing  chiefly  to  the  relatively  cool  cylinder 
walls,  but  to  some  extent  to  the  moisture  in  the  air  is  never  the  case. 

If,  however,  we  assume  an  exponent  of  1  1/3  or  4/3  we  shall  have  a 
fundamental  formula,  that  may  be  solved  by  the  more  familiar  opera- 
tions of  squaring,  cubing  and  extracting  the  square  and  cube  roots. 
We  shall  also  have  a  formula  that  will  give  results  approximating 
closer  to  practice  than  the  usual  formula,  because  it  allows  for  a  small 
loss  of  heat.  Absolute  pressures  and  temperatures  are  always  used. 

The  fundamental  formula  then  is 


(1) 
This  may  be  put  in  the  form, 

(2) 

which  is  a  more  convenient  form  to  use.     In  words  formula  No.  2  says: 
"The  pressure  multiplied  by  the  volume  and  this  product  by  the  cube 
root  of  the  volume  is  always  the  same,  that  is,  is  equal  to  a  constant  " 
Prom  equation  No.  1  the  following  formulas'  and  rules  may  be  de- 

24 


FIRST. 

To  calculate  the  pressure,  the  volume  being  known: 
Rule  1.     Divide  the  constant  by  the  volume  multiplied  by  its  own 
cube  root.  C 

That  is  (3)     p= 


We  generally  know  the  volume  at  the  beginning  of  compression, 
and  we  also  know  that  the  pressure  then  is  one  atmosphere,  say  14.7 
Ibs.  If  we  wish  to  know  the  pressure  at  some  other  volume,  we  may 
use  the  following  rule,  as  well  as  Rule  1. 

Rule  2.  Divide  the  greater  volume  by  the  lesser,  and  multiply  this 
quotient  by  its  own  cube  root.  Multiply  this  last  product  by  the  pres- 
sure at  the  larger  volume  and  we  have  the  pressure  required. 

That  is  v       /y~ 

(4)  P=P1^Vi 

For  example,  suppose  we  have  a  compressor  that  has  a  12"  stroke 
and  10  sq.  ins.  area  of  piston,  compressing  to  74.7  Ibs.  (60  Ibs.  gage). 
The  figure  shows  the  indicator  diagram.  For  convenience  we  measure 
the  volume  in  inches  of  cylinder  length. 

We  first  find  the  constant,  C,  by  formula  2.  At  the  commencement 
of  the  stroke  the  pressure,  P,  is  one  atmosphere  (14.7  Ibs.)  and  the 
volume  V  is  12"  of  cylinder  length.  Therefore,  by  formla  2 

14.7  X  12  X  f/r2  =  C. 

14.7  X  12  X  2.29  =  C  =  404. 

Now  suppose  we  want  to  know  what  the  pressure  is  after  the  piston 
has  traveled  6  inches.  The  volume  remaining,  or  the  distance  the 
piston  has  to  travel  yet  before  reaching  the  end  of  its  stroke,  is  six 
inches.  Therefore  by  Rule  1  we  divide  the  constant,  404,  by  the  vol- 
ume, 6",  multiplied  by  its  own  cube  root. 

404  -H  6  X  1^6  =404  -+-  (6  X  1.817)  =  404  -f- 10.9  =  37  Ibs. 
By  Rule  2,  we  first  divide  the  larger  volume  by  the  lesser,  12  -f-  6 
=  2. 

We  then  multiply  this  quotient  by  its  own  cube  root, 

2  X  1^2  =  2  X  1.26  =  2.52 

We  then  multiply  this  product,  2.52,  by  the  pressure  at  the  larger 
volume,  14.7,  thus  14.7  X  2.52  =  37  Ibs. 

Rule  2  is  the  simpler,  and  the  form  that  will  be  generally  used. 
We  notice  that  the  diagram  is  still  inaccurate  at  this  point. 


I      I      I     I      I      I      I      I      I      I  TT    I     I      I      i      I     i      I  -T-T 
ill       J\Q         \9         \f        \r        \6         Iff        U         la         ie         If 


SECOND. 

To  find  the  volume,  the  pressure  being  known. 

Rule  1  Divide  the  constant  by  the  pressure,  and  multiply  this 
quotient  by  its  own  square  root,  then  extract  the  square  root  of  the 
product,  that  is 

(5)     V= 

Rule  2  Divide  the  lesser  pressure  by  the  greater,  and  multiply  the 
quotient  by  its  own  square  root,  extract  the  square  root  of  this  product, 
and  multiply  the  larger  volume  by  this  square  root 


Thus,  if  we  want  to  know  the  volume  at  the  end  of  compression  in 
the  above  example,  when  we  know  the  pressure  is  74.7  Ibs. 

By  Rule  1.  We  divide  the  constant  by  the  pressure  404  -^  74.7  = 
5.408. 

We  then  multiply  this  quotient  (5.408)  by  its  own  square  root,  5.408 
X  1/O08  =  5.408  X  2.326  =  12.58. 

We  then  extract  the  square  root  of  this  product  (12.58)  \/ 12.58  = 
3.537  or  about  3.54  inches  of  cylinder  length. 

By  Rule  2.  We  first  divide  the  lesser  pressure  (14.7  Ibs.)  by  the 
greater  14.7  -=-  74.7  =  .1968. 

We  then  multiply  this  quotient  (.1968)  by  its  own  square  root,  .1968 
X  ]/. 1968  =  .1968  X  .444  =  .0874  and  extract  the  square  root  of  this 
product  (  N/M74)  v/T08T¥=.2956. 

We  then  multiply  the  greater  volume,  12",  by  this  root  (.2956),  12 
X  .2956  =  3.54"  inches  of  cylinder  length.  Which  is  the  result  re- 
quired. 


To  Find,  the  Net  Work  per  Stroke  of  the  Compressor. 

Rule.  First  multiply  the  stroke  (in  feet)  by  the  area  of  the  piston 
(in  square  inches)  and  this  product  by  58,  and  call  this  result  No.  1. 

Second.  Divide  the  greater  by  the  less  pressure,  extract  the  fourth 
root  (the  square  root  of  the  square  root)  of  this  quotient,  diminish  it 
by  unity  (1)  and  call  the  remainder,  result  No.  2. 

Third.  Multiply  result  No.  1  by  result  No.  2  and  the  product  is  the 
work  per  stroke  in  foot  pounds — (of  work  =  59  APV 


Thus,  if  we  wish  to  know  the  work  per  stroke  in  the  above  example, 
by  Rule  1,  we 

First.  Multiply  the  stroke  in  feet  (1)  by  the  area  of  the  piston  in 
square  inches  (10)  and  this  result  by  58,  so  that  we  have  1  X  10  X  58 
=  590.  Result  No.  1. 

Second.  We  divide  the  greater  (74.7)  by  the  less  pressure  (14.7) 
and  extract  the  fourth  root  of  the  quotient  and  diminish  it  by  one 

74.7  +  14.7  =  5.082;     4/57082  =  ^  /I/TOM  =  1/2^55  =  1.502 


Diminishing  this  by  one,  1.502  —  1  =  .502.     Result  No.  2. 

Multiplying  result  No.  1  by  result  No.  2  we  have  590  X  .502  =  296 
ft.  Ibs.  per  stroke. 

26 


The  above  rules,  and  formulas,  are  also  applicable  to  the  ammonia 
compressor. 


To  Find  the  Compression  Temperature. 

Rule  1.  Multiply  the  initial  temperature  (absolute)  by  the  fourth 
root  of  the  quotient  obtained  by  dividing  the  greater  by  the  less  pres- 
sure. The  result  is  the  final  temperature  absolute. 


Thus  in  the  example  taken,  assuming  that  the  initial  temperature  is 
60°  —  460  +  60  or  520  absolute,  we  have  ^=520  4/5.082=520  X  1.50 

=  781  absolute  or  781  —  .460  =  320°  by  the  thermometer. 

Compression  temperature  may  also  be  found  by  drawing  the  iso- 
thermal line  (steam  expansion  line)  under  the  actual  diagram. 

The  initial  absolute  temperature  is  to  the  final  absolute  temperature 
as  the  lengths  of  the  horizontal  lines  drawn  to  the  points  of  the  curves 
where  the  temperature  is  to  be  measured.  Thus  referring  to  the  figure 
or  diagram: 

T  :Tj  ::ab  :  ac 
ac  =  3.54 

14.7 
ab  =  -  X  12  =  2.36 

74.7 
.-.  2.36  :  3.51  ::  520  :  Tj  .-.  ^  =  780+ 


MISCELLANEOUS  RULES. 

The  work,  in  foot  pounds,  per  pound  of  air  compressed  may  be  ob- 
tained by  multiplying  the  rise  in  temperature  by  the  constant  213. 

The  weight  of  one  cubic  foot  of  air  in  pounds  at  any  temperature 
and  pressure  may  be  found  by  multiplying  its  pressure  by  the  constant 
2.707  and  dividing  by  its  absolute  temperature. 

Weight  per  cubic  foot  in  Ibs.  = 


Under  constant  pressure  the  volume  of  air  is  proportional  to  its 
absolute  temperature. 

Under  constant  volume  the  pressure  is  proportional  to  the  absolute 
temperature. 

The  specific  heat  of  air  is:  Under  constant  volume,  .1688;  under 
constant  pressure,  .2377. 

In  two  stage  compression  the  intermediate  pressure  is  equal  to  the 
square  root  of  the  initial  and  final  pressure. 


Question  1.  (1897-8.)  How  much  heat  is  produced  by  the  complete 
combustion  of  — (a)  one  pound  of  anthracite  coal,  (b)  one  pound  of 
bituminous  coal — of  average  quality?  Express  answer  in  heat  units 
and  also  in  foot  pounds.  Name  kind  of  coal 

(See  table  on  the  following  page) 

27 


Ans.  1. 
KIND  OF  COAL. 

Heat 

Evapo 
from  a 
21 

ration 
ndat 
2° 

||| 
£  ON 
§.««" 

0   «        "0 

lit! 

Value 

60$ 

m% 

5P 

£s  ° 
A 

PENNSYLVANIA  ANTHRACITE. 

12,000 
to 
14,900 
Average 
13,916 

7.45 
9.245 
8.63 

9.93 
12.32 
11.51 

10.07 

70 

j 

14,098 

8.74 

11.52 

13,551 

8.4 

11.  2L 

13,800 

8.56 

11.42 

11.17 

78 

13,104 

8.13 

10.84 

14,500 

9. 

12. 

11.13 

74 

i_/acKdw 

12  200 

7.57 

1009 

Honeybrook  Lehigh  
Massachusetts  Anthracite  

15,200 

9.44 

12.59 

10.75 

BITUMINOUS,  SEMI-BITUMINOUS,  ETC. 

R' 

14,000 

8.7 

11.6 

T  ^       ^Bl  '  'k 

14000 

8.7 

11.6 

14,400 

8.94 

11.93 

Pittsburgh  Coking  

14,400 
14  265 

8.94 
8.87 

11.93 
11.82 

Pocahontas,  run  of  mine  

14,289* 
13,614 

8.87 
8.46 

11.82 
11.27 

11.53 

8.74 

78 
62 

Pennsylvania,  semi-bituminous  

13,368 
13  000 

8.30 
807 

11.07 
10.76 

Bureau  Co     111                       .          

13,025 

8.07 

10.76 

11,083* 

6.88 

9.18 

7.78 

68 

12,139* 

7.54 

10.05 

8.25 

66 

12,240* 

7.60 

10.13 

8.98 

71 

Big  Muddy   Jackson  Co     111   

12  600 

7.83 

10.43 

Bituminous—  1896  tests—  heat  value  for 
ten  mines  scattered  over  Eastern  anc 
Southern   Ohio  (data  furnished  by 
Akron,  Ohio,  No.  28)  
Wood  (assumed  as  .4  of  coal_used)  .  .  . 
Lignite  

13,100 
10,300 

8.14 
3.48 
6.40 

10.85 
4.68 
8.50 

Crude  Petroleum  

19,200 

11.93 

15.90 

*1897 

Q.  2.  (1897-8.)  How  much  heat  will  be  produced  by  the  incomplete 
combustion  of  one  pound  of  coal  of  the  kind  assumed  in  Q.  1  (a)  ? 

Ans.  2.  Assuming  90%  carbon,  the  heat  produced  will  be  4,400  X 
.90  =  3,960. 


Q.  3.  (1897-8.)  What  are  the  products  of  combustion  under  the 
conditions;  of  Q.  1  (a)  and  what  are  their  approximate  proportions? 

Ans.  3.  Carbonic  acid  gas  (COa),  also  called  carbon  dioxide,  a 
colorless  gas  consisting  of  3/11  carbon  and  8/11  oxygen  by  weight; 
formed  by  the  union  of  the  oxygen  of  the  air  with  the  carbon  of  the 
coal  and  having  the  same  volume  as  the  oxygen  from  which  it  was 
formed:— steam  (H2O)  a  colorless  gas  formed  by  the  union  of  the 
oxygen  of  the  air  with  the  free  hydrogen  of  the  coal;  consisting  of  one 
part  by  weight  of  hydrogen  and  eight  parts  of  oxygen.  The  proportion 
of  steam  in  the  chimney  gases  is  usually  negligibly  small. 

28 


Q.  4.  (1897-8.)  What  are  the  products  of  combustion  under  the  con- 
ditions of  Q.  2,  and  what  are  their  approximate  proportions? 

Ans.  4.  Carbon  monoxide  (CO),  a  colorless  gas  consisting  of  3/7 
carbon  and  4/7  oxygen  by  weight,  formed  by  the  union  of  the  oxygen  of 
the  air  with  the  carbon  of  the  coal,  and  having  twice  the  volume  of 
the  oxygen  from  which  it  was  formed. 


Q.  5.  (1897-8.)  What  gas'  in  large  proportions  in  the  chimney  gases 
indicates  that  an  insufficient  quantity  of  air  is  being  supplied  to  the 
whole  or  a  part  of  the  fuel? 

Ans.  5.    Carbon  monoxide  (CO). 


Q.  6.  (1897-8.)  What  gas  in  large  proportions  in  the  chimney 
gases  indicates  that  too  much  air  is  being  supplied  to  the  furnace? 
What  proportion  of  this  gas  is  allowable  in  good  practice,  with  natural 
draft? 

Ans.  6.    Oxygen  (O2).     About  10%  by  volume. 


Q.  7.   (1897-8.)     What  would  be  the  weight  of  chimney  gases  per 
pound  of  coal  burnt,  in  good  practice,  with  natural  draft? 
Ans.  7.    About  25  Ibs. 


Q.  8.  (1897-8.)  What  is  the  approximate  specific  heat  of  chimney 
gases?  What  do  you  mean  by  this  answer? 

Ans.  8.  Twenty-four  one-hundredths  (.24).  Each  pound  of  the 
chimney  gases  raised  1°  F.  in  temperature,  absorbs  twenty-four  one- 
hundredths  (.24)  of  a  unit  of  heat. 


Q.  9.  (1897-8.)  With  the  facilities  of  the  average  engine  and  boiler- 
room  how  can  the  temperature  of  the  chimney  gases  be  approximately 
determined? 

Ans.  9.  (1)  By  a  thermometer,  properly  protected,  placed  in  the 
chimney  gases. 

(2)  By  exposing  a  certain  weight  of  iron  to  the  gases  until  it 
acquires  their  temperature  and  then  estimating  the  temperature  of  the 
iron  [see  answer    to  question  50   (1896-7).] 

(3)  By  the  melting  point  of  alloys  or  metals  exposed  to  the  gases. 


Q.  10.  (1897-8.)  What  proportion  of  the  heat  produced  by  the  burn- 
ing of  the  coal  should  go  to  the  formation  of  steam,  in  good  practice 
and  with  a  good  boiler  plant? 

Ans.  10.     From  60%  to  80%  (.60  to  .80). 


Q.  11.   (1897-8.)     How  can  the  amount  of  heat  produced  by  the  fur- 
nace be  approximately  estimated  and  how  can  the  proportion  of  said 

29 


number  of  pounds  of  the  same  bu rned  evaporated  by  the 

By  multiplying  the  numbf  °* J^"™    The  evaporation   is   usually 
heat   required   to   eyaporat e  one   poun^ d      1^ e         y  ^    ^ 

reduced   ^  an   equivalent   evapora^on    from  ^  ^  ^^  ^ 

Znd'of  waYer'at'^S'  to^e'pound  of  steam  at  the  same  temperature. 
-See  answers  to  last  year's  questions. 

n    12    ( 1897-8  )     What  would  be  the  weight  of  the  chimney  gases 
perQpound(of  coai^urnt,  in  good  practice,  with  a  forced  draft? 
Ans.  12.    About  19  Ibs. 

(CO,,  from  a  mixture 
(b)  For  the   absorption   of   free   oxygen    (O)    from   a  mixture   of 

Ans   13      (a)  A  solution  of  caustic  potash  (commercial)   in  water 
from  the  hydrant.     Proportions  about  1  lb.  of  potash  to  2  Ibs.  of  water. 

^fSffiSSySSto  acid  (pyrogallol)  in  water  mixed  with 
the  above  solution  of  caustic  potash.  Prof.  Thurston  used  5%  pyro- 
gallic  acid. 

Phosphorus  may  also  be  used. 


Q.  14.  (1897-8.)  Describe  a  cheap  and  simple  apparatus  by  which 
the  proportions,  by  volume,  of  carbonic  acid  gas  (CO2)  and  ot  tree 
oxygen  (O)  in  the  chimney  gases  may  be  determined.  Describe  meth- 
od of  use. 

Ans  14.  Ga.  No.  1  sends  the  following  description: 
"Fill  a  graduated  test-tube  with  chimney  gases.  Close  the  mouth, 
invert  the  tube  and  put  the  mouth  under  water.  Arrange  the  tube 
so  that  the  level  of  the  water  inside  and  out  is  the  same.  Now  intro- 
duce a  piece  of  caustic  potash  fastened  to  the  end  of  a  wire;  allow  it  to 
stand  about  45  mins.;  withdraw  the  potash.  It  will  be  found  that 
the  volume  (of  the  gas)  has  diminished,  which  represents  the  per  cent 
of  C02  in  gases.  Rearrange  the  tube  So  the  water  level  is  the  same 
inside  and  out;  introduce  a  piece  of  phosphorus  and  allow  to  stand  24 
hours.  Withdraw  the  phosphorus.  It  will  be  found  the  volume  has 
again  diminished,  which  represents  the  per  cent  of  oxygen." 

The  device  which  your  committee  has  constructed  and  used  is  de- 
scribed and  illustrated  in  a  separate  chapter. 


Q.  15.  (1897-8.)  Assuming  that  a  sufficient  amount  of  air  is  sup- 
plied to  the  fuel  and  that  the  per  cent  by  volume  of  carbonic  acid  (CO2) 
and  of  free  oxygen  (O)  in  the  chimney  gases  is  known,  how  can  the 
weight  of  the  chimney  gases  per  pound  of  coal  be  approximately  deter- 
mined? 

Ans.  15.  Add  the  per  cent  by  volume  of  carbonic  acid  (C02)  to  the 
per  cent  by  volume  of  oxygen  (02) ;  divide  this  result  by  the  per  cent 
by  volume  of  carbonic  acid,  and  multiply  the  quotient  by  the  constant 
number  ten  and  one-half  (10.5). 

39 


2.  Divide  the  constant  number  two  hundred  and  ten  (210)  by  the 
number  indicating  the  per  cent  by  volume  of  the  carbonic  acid  (C02). 

We  remark  that  the  above  rules  are  for  finding  the  weight  of  gases 
per  pound  of  coal,  and  not  per  pound  of  carbon  or  combustible  in  the 
coal.  It  is  therefore  necessary  to  assume  an  arbitrary  constant  which 
will  give  the  average  value.  For  the  combustible  the  above  constants 
would  be  about  12  and  240;  for  a  particularly  good  coal  they  would  be 
about  11  and  220. 


Q.  16.  (1897-8.)  Knowing  the  per  cent  by  volume  of  carbonic  acid 
gas  (CO-)  and  of  free  oxygen  (O)  in  the  chimney  gases,  how  can  the 
per  cent  by  volume  of  carbon  monoxide  (CO)  be  calculated,  neglecting 
water  vapor  and  hydrocarbons? 

Ans.  16.  Add  the  per  cent  by  volume  of  carbonic  acid  (COa)  and 
of  oxygen  (Oa),  multiply  the  result  of  5,  subtract  the  product  from  100 
and  divide  the  remainder  by  3. 


Q.  17.  (1897-8.)  If  coal,  having  a  heat  value  of  14,000  units  per  Ib. 
is  used,  the  feed-water  is  at  100  degrees,  the  pressure  is  70  Ibs.  gage,  it 
is  found  that  30  Ibs.  of  water  is  evaporated  by  4  Ibs.  of  coal  burned:  — 
What  per  cent  of  the  heat  produced  by  the  combustion  of  the  fuel  is 
used  in  making  steam? 

Ans.  17.  The  heat  produced  by  the  coal  should  be  14,000  X  4  = 
56,000.  The  factor  of  evaporation  for  feed-water  at  100°  and  pressure 
70  Ibs.  is  1.149.  Therefore  the  equivalent  evaporation  from  and  at  212° 
is  30  X  1.149,  which  is  34.47,  34.47  X  966  =  33,298,  which  is  the  number 
of  heat  units  used  in  making  steam.  Therefore  the  per  cent  of  the  heat 
used  in  making  steam  is  33,298  -=-  56,000  =  59.46. 


Q.  18.  (1897-8.)  If  under  the  conditions  of  Q.  17  it  were  found 
that  the  temperature  of  the  chimney  gases  was  450°  F.  above  the  air  in 
the  boiler  room,  and  that  40  Ibs.  of  air  was  being  supplied  per  pound 
of  coal  burned,  what  per  cent  of  the  heat  produced  by  the  combustion 
of  the  coal  would  be  going  out  of  the  chimney  with  the  gases? 

Ans.  18.  If  40  Ibs.  of  air  was  supplied  per  pound  of  coal  the  weight 
of  the  chimney  gases  would  be  about  41  Ibs.  per  pound  of  coal;  the 
heat  carried  up  the  chimney  by  this  would  be  about  41  X  450  X  .24  = 
4,428,  and  the  per  cent  is  4,428  ^  14,000  =  31.6  (about). 


Q.  19.  (1897-8.)  What  is  the  relative  approximate  permeability  to 
heat  of  clean  boiler  plate,  %"  thick,  and  the  same  plate  covered  with 
%"  of  hard  sulphate  scale? 

Ans.  19.     From  two  to  one  (2  to  1),  to  two  to  one  and  a  half  (2  to 

Q.  20.  (1897-8.)  How  is  the  condition  of  the  boiler  indicated  by  a 
change  in  the  temperature  of  the  chimney  gases,  and  why? 

Ans.  20.  A  rise  in  the  temperature  of  the  chimney  gases  would 
indicate  a  dirty  boiler,  as  scale  on  the  inside  or  soot  on  the  outside 
would  reduce  the  conductivity  of  the  metal,  and  a  greater  quantity  of 
heat  would  pass  out  with  the  gas'es. 


Q.  21.  (1897-8.)  A  horizontal  boiler  is  18  ft.  long,  72"  diameter, 
pressure  70  Ibs.  gage.  The  lower  end  of  a  twelve  inch  water  glass  is 
48"  from  the  lowest  point  of  the  boiler.  At  9  a.  m.  the  water  is  4" 


as  a 

of  steam  is  the  boiler  supplying  per  hour.' 


Ans  21  The  water  has  fallen  3"  or  one-quarter  of  a  foot.  The 
average  width  of  the  surface  while  falling  multiplied  by  the  length  of 
the  boiler  (18  ft.)  and  by  the  distance  through  which  the  level  has 
fallen,  gives  the  volume  of  the  water  used.  This  expressed  in  cubic 
feet  and  multiplied  by  56.8,  the  weight  of  water  at  the  temperature 
corresponding  to  70  Ibs.  gage  (316°),  will  give  the  weight  of  water- 
evaporated  in  the  given  time—  three-quarters  of  an  hour.  This  result 
divided  by  .75  will  give  the  weight  of  water  per  hour  that  the  boiler 
is  evaporating. 

To  find  the  average  width  of  the  surface  one  might  take  the  width 
at  the  higher  and  at  the  lower  levels,  add  the  two  together  and  divide 

The  width  of  the  surface  (a  d  Fig.  1)  at  the  higher  level  may  be 
found  in  subtracting  the  square  of  its  height  (be)  above  the  center 
of  the  boiler  from  the  square  of  the  radius  (36"),  extracting  the  square 
root  of  the  result  and  multiplying  by  2. 

Thus:  —  Radius  equals  36",  square  of  radius  =  1296.  Height  from 
center  of  boiler  20",  square  of  height  400;  1296  —  400  =  896.  The 
square  root  of  896  is  29.933,  and  twice  this  is  59.866,  which  is  the 
width  of  the  surface  at  the  higher  level.  The  width  of  the  surface  at 
the  lower  level  may  be  found  in  the  same  way  to  be  63.46;  59.866  + 
63.46  -T-  2  =  61.66",  or  about  5.14  ft,  as  the  average  width. 

It  would  perhaps  be  better  to  draw  a  diagram  like  Fig.  1  to  as  large 
a  scale  as  practicable,  say  one-half,  then  draw  a  line  f  g  half-way  be- 
tween the  lines  a  d  and  h  e,  which  indicate  the  higher  and  lower  levels. 
Then  carefully  measure  the  line  g  f.  With  the  scale  suggested,  the  line 
will  be  found  to  be  about  30  13/16  long,  which,  multiplied  by  2,  because 
of  the  reduced  scale,  will  give  about  61.63". 

This  illustrates  the  simplicity  and  practical  accuracy  of  graphical 
methods  of  calculation. 

Taking  the  average  width  as  5.14  ft,  the  length  as  18  ft.  and  depth 
as  .25  ft.,  the  volume  of  the  water  evaporated  would  be  5.14  X  18  X 
.25  =  23.13  cubic  feet,  and  its  weight  is  23.13  X  56.8  =  about  1313  Ibs. 
This  multiplied  by  4  and  divided  by  3  gives  about  1751  Ibs.  of  water  per 
hour. 


32 


Q.  25.  (1897-8.)  The  gage  pressure  being  120  pounds,  the  feed 
water  110°  and  the  engine  using  15  Ibs.  of  steam  per  horse-power-hour, 
what  part  of  the  heat  that  went  to  the  making  of  steam  is  transformed 
into  useful  work  by  the  engine?  What  per  cent  is  this  of  the  heat 
generated  in  the  furnace,  the  efficiency  of  the  furnace  and  boiler  being 
.70? 

Ans.  25.  (a)  The  total  heat  in  steam  above  32°  due  to  120  Ibs. 
gage  pressure  is  1188  H.U.  and  the  heat  already  in  the  water  is  110  — 
32  =  78  H.U.  Then,  the  heat  required  to  change  1  Ib.  water  at  110° 
into  steam  at  120  Ibs.  gage  pressure  is  1188  —  78  =  1110  H.U.  The  engine 
in  question  uses  15  Ibs.  of  steam  per  horse-power,  which  is  equal  to 
15  X  1110  =  16,650  H.U.  As  1,980,000  is  equal  to  one  horse-power  per 
hour,  and  1  H.U.  equals  778  ft.  Ibs.  of  work. 
1,980,000 

=  2545. 

778 

the  H.U.  changed  into  work  per  HP.  per  hour.  Then  the  per  cent  of 
heat  that  went  to  make  steam  and  was  changed  into  work  by  the  en- 
gine is 

2545 

—  X  100  =  15.22%. 
16650 

(b)     If  the  furnace  and  boiler  has  an  efficiency  of  70%  and  the 
boiler  uses  16,650  H.U.  for  15  Ibs.  of  steam,  the  heat  of  the  furnace 
would  have  to  produce  for  each  15  Ibs.  of  steam 
16650 

—  =23785.7  H.U. 
.70 

and  the  per  cent  of  this  heat  that  went  to  do  useful  work  is 
2545 

X  100  =  10.7. 

23785.7 


Q.  42.  (1897-8.)  Give  three  ways  of  calculating  the  area  of  the  end 
of  the  boiler  above  the  tubes,  which  is  supported  by  braces? 

Ans.  42.  Taking  the  segment  as  less  than  a  half  circle  and  of  the 
size  that  it  is  usually  necessary  to  brace  on  the  end  of  a  boiler: 

FIRST  METHOD. 

Take  one-half  the  area  of  the  entire  circle  of  which  the  segment  is  a 
part.  Call  this  result  No.  1.  Multiply  the  diameter  of  said  circle  by  the 
height  of  the  base  of  the  segment  above  the  center  of  the  circle.  Call 
this  result  No.  2.  Subtract  result  No.  2  from  result  No.  1  and  the 
remainder  is  the  area  of  the  segment  very  nearly.  The  result  is  a  little 
too  small. 

SECOND   METHOD. 

Subtract  twice  the  distance  from  the  base  of  the  segment  to  the  cen- 
ter of  the  circle  from  half  the  circumference  of  the  circle  and  multiply 
this  result  by  one-half  the  radius  of  the  circle.  Call  this  result  No.  1. 
Multiply  the  radius  of  the  circle  by  the  height  of  the  base  of  the  seg- 
ment above  the  center.  Call  this  result  No.  2.  Subtract  result  No.  2 
from  result  No.  1  and  the  remainder  is  the  area  of  the  segment  very 
nearly.  The  result  is  a  little  too  small. 

THIRD  METHOD. 

Draw  the  segment  to  as  large  a  scale  as  convenient  and  measure  its 
area  by  the  planimeter. 

Inasmuch  as  two  or  more  methods  of  solving  a  problem  are  some- 
times useful  as  a  check  upon  each  other,  and  as  some  will  prefer  to  use 

33 


one  way  and  some  another,  the  graphical  methods  of  solving  the  seg- 
ment problem  might  be  entertaining  and  useful.  We  believe  that  these 
methods  as  given  are  sufficiently  accurate  for  practical  purposes.  The 
second  method  is  quite  accurate  if  one-half  the  base  of  the  segment  is 
taken  instead  ot  the  radius  of  the  circle,  in  calculating  the  triangular 
area. 

AN  EXAMPLE. 

The  circle  is  53"  in  diameter.  The  height  of  the  base  of  the  segment 
above  the  center  of  the  circle  is  7.5".  (See  Fig.  6.) 

First  Method: — The  area  of  the  circle,  a  b  c  f,  53"  diameter,  is 
2206.18  sq.  ins.  The  area  of  the  semi-circle,  g  b  h,  is  one-half  of  2206.18 
or  1103.09  sq.  ins.  Which  is  result  No.  1. 

If  you  multiply  the  diameter  g  h,  equal  53",  by  the  height  e  d,  equal 
7.5",  you  get  397.5  sq.  ins.,  which  is  a  little  greater  than  the  area  g  a  c  h. 
If  you  take  the  area  g  a  c  h  from  the  area  g  a  b  c  h,  you  have  left  the 
area  of  the  segment,  a  b  c  e.  Therefore  1103.09  —  397.5  =  705.59  sq.  ing., 
the  approximate  area  of  the  segment. 

Second  Method: — If  you  subtract  twice  the  height  e  d,  from  the  semi- 
circumference  g  a  b  c  h,  you  have  approximately  the  length  of  the  line 
a  b  c.  If  you  multiply  the  line  a  b  c  by  one-half  the  radius,  a  d,  you 
have  the  area  d  a  b  c.  If  you  multiply  the  radius  by  the  height  e  d, 
you  have  approximately  (a  little  large)  the  area  of  the  triangle  dace. 
If  you  take  the  area  dace  from  the  area  d  a  b  c,  you  have  the  area 
a  b  c  e  remaining,  which  is  the  area  of  the  segment. 


FIG.  6 

Referring  again  to  Fig.  6:— The  circumference  of  a  circle  having  a 

diameter  of  53"  is  166.5".    One-half  this  (the  line  g  b  h),  is  83.25".     If 

from  this  you  subtract  twice  the  height  d  e,  that  is  twice  7.5"  or  15", 

you  have  83.25  — 15  =  68.25",  which  is  approximately  the  length  of  the 

"lie_a  b  c-      [f  you  multiply  this  by  one-half  the   radius,   you   have 

J.25  =  904.31  sq.  ins.,  or  the  area  d  a  b  c.    This  is  result  No.  1. 

7  K  0       J  multiply  the  height  d  e,  by  the  radius  of  the  circle,  you  have 

198.75  sq.  ins.,  which  is  nearly  the  area  of  the  triangle  a  c 

d.    A  little  large.    This  is  result  No   2 

—  7^ubtJactinS  result  No'  2  from  result  No-  x  y°u  have  904.31—198.75 
56  sq.  ink,  for  the  area  of  the  segment,  which  is  a  little  small 
because  the  area  of  the  triangle  was  a  little  large. 

prmula  of  the  answer  to  last  year's  Q.  No.  15,  and  by  accur- 
710  l£  iS"ng  SeC°nd  method'  tne  area  wil1  be  found  to  be  about 


34 


Q.  43.  (1897-8.)  If  a  boiler  brace  is  attached  to  the  shell  4  ft.  from 
the  end  of  the  boiler  and  to  the  end  of  the  boiler  2  ft.  from  the  shell, 
what  would  be  the  strain  upon  the  brace  as  compared  to  an  end-to-end 
brace  supporting  the  same  area? 

Ans.  43.  The  strain  upon  the  brace  attached  to  the  shell  would  be 
to  the  strain  upon  the  end-to-end  brace  as  the  length  of  the  first  men- 
tioned brace  is  to  the  distance  of  its  point  of  attachment  to  the  shell 
from  the  end  of  the  boiler.  In  this  instance  1.118. 


Q.  48.  (1897-8.)  What  are  the  different  ways  in  which  a  riveted 
joint  may  give  way? 

Ans.  48.  The  plate  may  tear  across  along  th'e  line  of  least  cross- 
section  a,  b,  Fig.  7,  (a). 

2.  The  plate  and  rivet  may  be  crushed,  as  shown  in  (b). 

3.  The  plate  may  break  across  in  front  of  the  rivet,  (c). 

4.  The  rivet  may  shear  across,   (d). 


Q.  49.  (1897-8.)  What  is  the  general  rule  for  selecting  the  diameter 
of  rivets  for  a  given  thickness  of  boiler  plate?  What  should  be  consid- 
ered in  determining  the  diameter  of  rivets? 

Ans.  49.  If  the  rivet  holes  are  to  be  punched,  the  punch  should 
have  a  diameter  as  great  as  the  thickness  of  the  plate,  otherwise  it  is 
liable  to  be  broken.  Drilled  holes  are  not  made  less  in  diameter  than 
the  thickness  of  the  plate. 

As  the  shearing  strength  of  the  rivet  increases  with  the  square  of 
its  diameter,  and  the  crushing  strength  of  the  plate  in  front  of  the 
rivet  increases  only  as  the  first  power  of  the  diameter,  there  will 
evidently  soon  come  a  time,  as  the  rivet  is  increased  in  diameter,  when 
the  shearing  strength  of  the  rivet  is  greater  than  the  crushing  strength 
of  the  plate.  A  correctly  designed  joint  should  be  equally  strong  in  all 
its  parts. 

The  rivets  should  be  close  enough  together  to  hold  the  joint  tight. 

The  rule  given  by  Unwin  is  to  make  the  diameter  of  the  rivet  1.2 
times  the  square  root  of  the  thickness  of  the  plate — 
d  =  1.2  Vt. 


-0- 


(b) 


(c) 


Q.  50.  (1897-8.)  What  effect'  does  the  length  of  a  tube  have  on  its 
collapsing  strength? 

Ans.  50.    No.  12  of  Boston,  Mass.,  says: 

"The  longer  the  tube  the  lower  will  be  its  collapsing  pressure.  First, 
because  a  short  tube  will  retain  its  circular  form  better  than  a  long 
one  owing  to  the  tendency  of  the  long  tube  to  sag  in  the  middle  and 

35 


get  out  of  shape  and,  again,  the  short  tube  has  the  advantage  of  the 
support  of  the  heads  to  which  it  is  attached.  This  support  would  also 
be  given  to  the  long  tubes,  but  would  not  have  the  same  effect  at  the 
middle  of  its  length.  Large  flues,  like  those  in  marine  boilers  are 
strengthened  by  rings  placed  at  regular  intervals  to  hold  the  flue  in 
shape.  Another  method  is  to  make  the  flue  of  short  lengths,  joining  the 
ends  by  riveting  them  to  the  rings.  Another  method  is  to  make  the 
flues  corrugated.  All  of  which  goes  to  show  that  a  short  tube  will  stand 
a  greater  external  pressure  than  a  long  one." 

While  the  above  seems  about  right  to  the  committee,  still  it  is  said 
that  for  the  purpose  of  calculating  the  thickness  of  the  tube,  the  length 
beyond  ten  or  twelve  diameters  may  be  neglected. 


Q.  58.  (1897-8.)  If  a  boiler  weighing,  with  its  contained  water, 
10  tons,  is  supported  by  four  symmetrically-placed  round  wrought  iron 
rods,  what  should  be  the  diameter  of  the  rods? 

Ans.  58.  Each  rod  would  have  to  support  %  of  the  entire  weight 
(10  tons),  or  5000  Ibs.  Allowing  10000  Ibs.  as  the  safe  working  strain 
per  square  inch  of  cross-section  each  rod  would  require  .5  of  a  square 
inch  cross-section,  which  corresponds  to  a  diameter  of  .8". 

To  allow  for  the  change  of  temperature  and  corrosive  action  to  which 
the  rods  would  be  subjected,  the  rods  would  probably  best  be  made  1" 
or  1.125"  in  diameter. 


BOOKS  REFERRED  TO  BY  THE  COMMITTEE  OF  1897-8. 

Gas  and  Fuel  Analysis,  for  Engineers—  Gill  —  published  by  John 
Wiley  &  Son. 

Hempel's  Gas  Analysis—  translated  by  Dennis  —  published  by  Mac- 
millan  &  Co. 

Boiler  Tests—  by  Geo.  H.  Barrus—  published  by  the  author. 

We  would  refer  any  one  wishing  to  read  up  on  the  subject  of  the 
graphical  representation  of  the  forces  of  inertia  in  steam  engines,  to 
Holmes  on  "The  Steam  Engine,"  published  in  Appleton's  "Text  Books 
of  bcience  series.  The  subject  is  treated,  however,  in  nearly  all  books 
on  mechanics  or  the  steam  engine. 

Upon  the  subject  of  lubricating  oils.  Thurston's  work  on  "Lost  Work 
^Machinery,"  etc.,  published  by  Wiley,  or  G.  H.  Hurst  on  "Lubricating 

Gill  on  Oil  Testing. 

*"*  standard  and  reliable  works  on  the  strength  of 


Strains  in  Framed  Structures,  by  Bindon  B.  Stoney 
Unwm's  Machine  Design. 

Rauleaux'g  Constructor,  translated  by  Supplee 
Weisbach's  Mechanics  Vol.  1,  translated  by  Cox 

!f  MQthe  TFk»  ±Silvanus  R  Thompson,  "Elements  of  Elec- 
erv"  ™*  «Mvgn  S£V  "Electromagnets  and  Electromagnetic  Machin- 
thls  subject  1C  Machinery>"  are  ^out  as  good  as  any  on 

iroi£  rZkrn?£titled  "M,ec*ai»cal  Draft,"  published  as  an  advertisement, 
the  best  wort     ref£rd  ^  flrst  hundred  or  hundred  and  fifty  pages  as 
st  work  on  the  subject  of  combustion  they  have  met. 

The  Educational  Committee. 

INTRODUCTION  TO  QUESTIONS  83  TO  112    (1898-9). 
Tnt^d  loV^f  th  ^S  been  V™*a™*  with  care.    The  questions 

36 


ulty  of  as  much  value  as  the  more  refined  calculations  which  the  same 
points  are  also  susceptible  of — neither  is  superfluous  and  both  should 
be  cultivated  energetically. 


Q.  83.  (1898-9.)  At  what  standard  of  temperature  is  the  evapora- 
tion of  steam  boilers  usually  expressed,  irrespective  of  the  actual  tem- 
perature of  the  feed  water  or  the  pressure  at  which  the  steam  is 
taken  off? 

Ans.  83.  For  comparison  it  is  usual  to  reduce  the  results  of  boiler 
tests  to  a  known  standard.  This  standard  is  technically  known  as  the 
"equivalent  evaporation  from  and  at  212°."  This  means  that  the  evap- 
oration is  considered  to  have  taken  place  at  mean  atmospheric  pressure 
and  at  a  temperature  due  to  that  pressure  and  the  feed  water  also  being 
assumed  as  supplied  at  that  temperature. 

One  pound  of  water  evaporated  "from  and  at  212°"  is  equivalent  to 
965.7  B.  T.  units.  

Q.  84.  (1898-9.)  Explain  the  difference  between  a  pound  of  coal  and 
a  similar  weight  of  "combustible." 

Ans.  84.  The  distinction  between  equal  weights  of  coal  and  com- 
bustible is  as  follows:  A  pound  of  coal  always  contains  a  "per  cent" 
of  moisture  and  non-combustible  matter  in  the  form  of  ash  and  clinker. 
After  deducting  this  the  remainder  is  useful  for  combustion.  A  pound 
of  combustible  consists  of  such  substances  as  can  be  completely  burned 
or  consumed — leaving  no  refuse. 

Q.  85.  (1898-9.)  How  many  pounds  of  coal  burned  each  hour,  per 
square  foot  of  grate,  is  considered  a  fair  and  economical  rate  of  com- 
bustion? 

Ans.  85.  When  economy  is  a  consideration  the  rate  of  combustion 
must  evidently  not  exceed  the  absorbing  capacity  of  the  boiler.  With 
grate  and  heating  surfaces  properly  proportioned  the  best  economy  is 
claimed  for  a  rapid  combustion  of  the  fuel. 

From  15  to  20  Ibs.  of  coal  per  hour  burned  on  each  square  foot  of 
grate  is  considered  good  practice  for  factory  boilers',  the  latter  rate 
requires  an  exceptionally  strong  draft  and  is  well  up  to  the  limit  unless 
mechanical  means  are  resorted  to  for  hastening  the  consumption  of  the 
fuel. 

Q.  86.  (1898-9.)  What  is  a  good  "evaporation"  per  pound  of  coal  for 
boilers  of  horizontal  tubular  type,  under  fair  average  conditions? 

Ans.  86.  The  evaporation  from  boilers  of  the  horizontal  tubular 
type,  under  fair  conditions,  may  range  from  7%  to  9*£  Ibs.  The  lower 
figure  is  probably  more  common  and  the  latter  may  be  slightly  exceeded. 
Much  depends  upon  the  kind  of  coal  and  the  methods  and  manner  in 
which  the  plant  is  conducted. 


Q.  87.  (1898-9.)  What  "check"  or  test  is  necessary  to  obviate  decep- 
tive results  or  magnify  the  evaporative  efficiency  of  a  steam  boiler? 

Ans.  87.  A  calorimeter  test  should  always  be  made  where  accurate 
results  are  wanted  to  determine  the  quality  of  steam  and  per  cent 
of  primage.  This  percentage  must  be  deducted  from  the  apparent  evap- 
oration to  arrive  at  a  true  result;  that  is,  actual  evaporation. 


Q.  88.  (1898-9.)  As  an  example,  name  a  rate  of  evaporation,  per 
pound  of  combustible,  that  you  would  regard  with  suspicion. 

Ans.  88.  Efficiency  of  the  furnace  and  the  boiler  is  determined  by  a 
comparison  of  the  heat  units  accounted  for  in  the  steam  produced  and 
the  theoretical  value  of  the  fuel  consumed.  Combustion  is  always  more 

37 


or  less  incomplete;  fuel  is  lost  by  dropping  through  the  grate;  heat  is 
carried  away  by  the  hot  gases  and  lost  by  radiation,  hence  when  the 
difference  between  the  theoretical  and  realized  values  seems  too  small 
to  cover  these  losses,  there  is  good  cause  for  caution  in  accepting 
the  accuracy  of  figures  which  represent  an  abnormal  rate  of  evaporation. 
-  Naming  12  Ibs.  of  water  evaporated  from  and  at  212°  Fahr.  as  a  rate 
open  to  suspicion  we  may  reckon  966  X  12  =  11592  B.T.U.  accounted  for 
in  steam.  This  is  about  82%  of  14000,  the  latter  figure  being  the 
assumed  value  of  the  combustible.  Experience  shows  that  18%  will 
not  cover  the  losses  attending  common  practice;  40%  is  not  an  unusual 
loss  and  less  than  20%  is1  seldom,  if  ever,  attained. 


Q.  89.  (1898-9.)  For  a  safe  working  pressure  of  100  Ibs.  gage  pres- 
sure give  the  maximum  diameter  and  also  the  greatest  length  desirable 
for  horizontal  tubular  boilers. 

Ans.  89.  A  well-proportioned  horizontal  externally  fired  tubular 
boiler  working  under  100  Ibs.  pressure  should  not  be  larger  than  72" 
diameter  and  18  feet  long. 

The  Hartford  Steam  Boiler  Inspection  and  Insurance  Company,  at 
the  request  of  this  committee,  kindly  furnished  the  following  on  this 
question : 

"This  company  does  not  approve  of  the  employment  of  plate  thicker 
than  y2"  in  the  construction  of  externally  fired  boilers.  Keeping  within 
this  limit  of  thickness  and  assuming  the  tensile  strength  of  plate  at 
60000  and  that  all  the  usual  requirements  as  to  ductility  and  elongation 
are  complied  with.  The  diameters  for  a  plate  thickness  of  15/32"  are 
as  follows,  the  sizes  varying  with  the  efficiency  of  the  joint  used  in  the 
longitudinal  seams: 

Safe  working  pressure  100  Ibs.     Factor  of  safety — five. 

(a)  For  double-riveted  lap-joint  having  70%  efficiency,  diameter  78". 
Proof: 

60000  X  0.46875  X  .70        19687.5 

—  =  —       —  =  101  Ibs. 
(Radius)  39  X  5  195 

(b)  For  triple-riveted  lap-joint  having  75%  efficiency  the  diameter 
may  be  increased  to  84  inches. 

Proof: 

60000  X.  0.46875  X  75   21094 

—  —  = =  100  Ibs. 

(Radius)  42  X  5      210 

(c)  For  triple-riveted  butt  joint  with  double  straps  and  securing 
an  efficiency  of  86%,  the  corresponding  diameter  is  96  inches. 

Proof: 

60000  X  0.46875  X  .86        24187.5 

—  =  100%  Ibs. 
(Radius)  48  X  5  240 

Length  of  boiler  having  3V2"  or  4"  tubes  should  not  exceed  18  feet. 
The  proper  length  should  be  determined  by  the  ratio  of  heating  surface 
to  grate  area,  and  the  area  of  tube  opening  to  grate  area.  By  our  tests 
we  have  found  the  former  should  be  about  46  to  1  and  the  latter  when 
bituminous  coal  is  used  should  have  an  area  of  from  1/6  to  1/7  of  the 
grate  surface. 

All  specifications  issued  by  this  company  for  this  type  of  boiler  have 
a  factor  of  safety  of  five  and  is  considered  none  too  much 

The  efficiency  of  the  three  joints  mentioned,  that  is,  70,  75  and  86% 
%£?££*  P^te  when*  properly  proportioned,  can  be  realized.  The 
latter  form  of  joint  is  illustrated  herewith  " 


WfL&t9,?h    (1898;9-)T,  ^hat  is  £enerally  claimed  for  all  boilers  of  the 
water-tube  class?     Define  also  the  advantages  and  the  disadvantages 


ANSWER  No.  89. 
39 


attributed  to  those  constructed  with  straight,  and  with  curved,  water 

tubss? 

An's  90     It  is  generally  claimed  for  water-tube  boilers: 

(a)     They  are  safe  from  destructive  explosion  at  extremely  high 

pressures^  ^  forced  far  bey0nd  their  normal  capacity  with  impunity. 

(c)  Occupy  small  space  and  weigh  less  per  unit  of  power  developed. 

(d)  Economy  in  the  use  of  fuel. 

(e)  Are  easily  cared  for,  etc.,  etc. 

Some  of  these  claims,  in  some  instances  at  least,  have  no  founda- 
tion in  fact. 

The  advantages'  of  straight  water  tubes  are,  that  they  can  be  easily 
inspected  and  cleaned  and  a  standard  tube  of  the  required  length  can 
be  used  for  renewals.  One  alleged  disadvantage  is,  that  owing  to 
their  straight  form  they  are  liable,  on  account  of  expansion  and  con- 
traction, to  severely  strain  the  headers  they  are  expanded  into.  This, 
however,  is  thought  to  be  more  imaginary  than  real  and  the  thousands 
of  boilers  of  this  type  in  successful  daily  use  bear  out  this  assumption. 

The  advantages  claimed  for  a  "curved"  tube  are,  that  neither  the 
tube  or  its  connections  are  injured  by  unequal  expansion  or  contraction. 
Such  tubes  are,  however,  difficult  to  properly  inspect  and  clean;  they 
are  hard  to  repair,  and  this  is  especially  true  of  boilers  containing 
different  lengths  and  bends.  There  are  unusual  conditions  where  boilers 
with  bent  tubes  are  a  necessity  and  are  also  desirable,  but  we  do  not 
believe  that  these  conditions  obtain  in  stationary  practice.  There  are 
many  of  this  type  and  most  of  them  are  a  delusion  and  a  snare  in  bad 
water. 

Q.  91.  (1898-9.)  How  many  square  feet  of  heating  surface  per 
horse-power  is  a  customary  allowance  in  horizontal  tubular  boilers? 
Give  also  a  similar  approximation  for  water-tube  boilers  of  average 
efficiency? 

Ans.  91.  Fifteen  square  feet  of  heating  surface  per  horse-power  is  a 
common  allowance  for  externally  fired  boilers  of  the  horizontal  tubular 
type,  and  ordinarily  this  is  sufficient  when  such  boilers  are  used  to  sup- 
ply simple  throttling  engines.  For  "Corliss"  class  of  engines  the  allow- 
ance is  more  usually  restricted  to  12  square  feet. 

Water  tube  boilers  are  rated  at  7  to  11  sq.  feet.  All  such  ratings 
should  be  regarded  only  as  an  approximation;  a  better  method  is  to 
take  into  consideration  the  various  condition's  which  apply  to  any  par- 
ticular case. 

Q.  92.  (1898-9.)  Give  an  average  ratio  of  grate  and  heating  surfaces 
in  horizontal  tubular  boilers  which  give  good  results  and  note  such 
conditions  as  may  prompt  a  change  in  the  proportions  given? 

Ans.  92.  For  horizontal  tubular  boilers  a  ratio  of  heating  surface 
approaching  35  to  46  to  one  square  foot  of  grate  is  usually  productive 
of  good  results. 

High  furnace  temperature  and  rapid  combustion  yield  economy;  the 
rate  of  combustion  is  always  determined  by  the  "draft"  and  the  more 
rapid  the  consumption  of  fuel  on  the  grate,  the  greater  should  be  the 
extent  of  the  heating  surface.  Draft  in  turn,  however,  is  ordinarily 
dependent  on  the  temperature  of  the  escaping  gases,  hence  the  increased 
ratio  referred  to  must  be  limited,  of  course,  to  a  point  where  the  draft 
remains  equal  to  the  requirements  of  the  furnace. 


Q.  93.  (1898-9.)  For  a  plant  requiring  200  HP,  working  steam 
pressure  100  Ibs;.,  and  which  is  to  be  operated  reasonably  free  from 
interruption,  due  to  cleaning  and  repairs,  give  general  dimensions  and 
the  number  of  horizontal  tubular  boilers  you  would  install.  State  also 


how  such  should  be  set,  i.  e.,  singly  or  in  pairs,  to  fill  the  requirements 
at  a  minimum  cost. 

Ans.  93.  (a)  Assuming  that  the  200  HP  plant  referred  to  in  the 
question  is  to  be  used  for  manufacturing  purposes  and  is  not  to  run  on 
Sundays  or  holidays  the  boilers  required  would  be  two  66"  dia.  16  ft. 
horizontal  tubular  type — 54-4"— tubes.  To  be  of  standard  design  and 
material,  first-class  workmanship  and  capable  of  safely  carrying  100  Ibs. 
pressure  per  square  inch.  Boilers  to  be  set  over  separate  furnaces  and 
properly  connected  so  they  could  be  run  independently  if  necessary. 

(b)  Another  answer  provides  a  third  boiler — which  would  be  neces- 
sary, of  course,  for  a  plant  designed  for  continuous  operation.  This 
answer  is  expressed  as  follows: 

We  would  recommend  three  boilers  of  the  horizontal  tubular  type 
set  singly;  that  is,  with  partition  walls  between  adjacent  boilers — each 
to  be  provided  with  separate  grate,  steam  and  water  connections, 
enabling  any  one  of  the  three  to  be  operated  independently.  Bach 
smoke  connection  to  stack  to  have  damper  and  each  boiler  to  be  pro- 
vided with  an  independent  gage  in  addition  to  the  steam  gage  con- 
nected to  the  main  header. 

Each  boiler  to  develop  100  HP;  that  is,  must  be  capable  of  evaporat- 
ing 3450  Ibs.  of  water  from  and  at  212°  per  hour.  Allowing  12  square 
feet  of  heating  surface  per  HP  would  therefore  require  1200  square 
feet  in  each,  boiler. 

A  boiler  66"  diameter— 18  feet  long,  fitted  with  54-4"  tubes  fulfills 
this  requirement. 

Q.  94.  (1898-9.)  Describe  briefly,  without  detail,  the  method  of 
supporting  horizontal  tubular  boilers  which  you  regard  with  favor? 

Ans.  94.  Boilers  are  best  supported  by  being  hung  up  on  steel  fram- 
ing in  a  way  that  the  brick  work  is  relieved  from  all  weight.  This 
makes  it  much  easier  on  their  "settings"  and  greatly  facilitates  work  in 
case  of  needed  repairs.  Properly  designed,  such  an  arrangement  allows 
freedom  for  expansion,  without  cracking  brickwork. 


Q.  95.  (1898-9.)  Which  is  preferable  in  horizontal  tubular  boilers, 
the  dome,  steam-drum,  or  dry-pipe? 

Ans.  95.  The  dome  or  steam  drums  are  expensive  adjuncts  to  hori- 
zontal tubular  boilers  and  are  useless  on  boilers  which  are  well  designed 
and  proportioned  with  a  view  of  dispensing  with  either.  A  properly 
designed  "dry-pipe"  is  an  equivalent,  answering  every  requirement 
when  all  other  things  are  equal. 


Q.  96.  (1898-9.)  Do  you  consider  the  advantages  claimed  for  mud- 
drums  sufficient  to  outweigh  such  objections  as  their  use  entails? 

Ans.  96.  The  trend  of  modern  "ways"  is  to  dispense  with  mud  drums 
on  horizontal  tubular  boilers.  They  deteriorate"  rapidly  and  are  there- 
fore considered  costly  and  not  ordinarily  necessary,  although  in  Missis- 
sippi River  practice  and  other  places  where  only  muddy  unfiltered  water 
is  available,  the  mud-drum  is  still  regarded  with  some  favor. 


Q.  97.  (1898-9.)  What  style  of  cock  or  valve  is  best  adapted  for 
lower  "blow-off"  of  boilers? 

Ans.  97.  For  "blow-off"  use  a  valve  with  passages'  calculated  to  pass 
freely  all  particles  likely  to  get  into  the  blow-off  pipe.  A  globe  valve  is 
out  of  place,  because  it  does  not  meet  the  above  requirement.  Valves 
"handle"  easier  than  plug  cocks  and  are  less  liable  to  "jar"  and  thereby 
injure  the  pipe  and  connections.  Asbestos-packed  cocks  are  claimed  to 
be  more  satisfactory  than  the  ordinary  kind. 


Q  5,8.  (1898-9.)  Give  the  proper  position  for  a  "surface  blow"; 
state  how  its  details  should  be  arranged  and  explain  the  purpose  it 
serves. 

Ans  98  A  "surface  blow-off,"  as  applied  to  a  horizontal  tubular 
boiler  consists  of  a  flat  funnel-shaped  appendage  inside  of  the  shell, 
preferably  located  near  the  rear  end  with  the  wide  mouth  of  the  funnel 
placed  at  the  water  line  and  directed  toward  the  front  of  the  boiler. 
This  should  be  connected  to  a  1%"  pipe  leading  out  of  the  back  head  of 
boiler  and  a  valve  should  be  located  in  an  accessible  position  so  it  can 
be  used  frequently  while  the  boiler  is  in  operation. 

The  purpose  served  is  the  removal  of  the  sludge  and  scum  that  is 
always  more  or  less  upon  the  surface  of  the  water  when  boilers  are 
being  worked.  When  properly  installed  and  attended  to,  a  surface 
blow-off  is  very  helpful  and  greatly  relieves  certain  conditions  which 
tend  to  cause  "foaming." 


Q.  99.  (1898-9.)  What  form  of  connection  do  you  approve  for  a 
water-column  and  how  should  gage-cocks  and  water-glass  be  placed 
with  reference  to  the  tubes  in  horizontal  tubular  boilers? 

Ans.  99.  A  water-column  should  be  connected  to  boiler  with  pipe  not 
smaller  than  l1^".  (The  Hartford  Steam  Boiler  Inspection  and  Insur- 
ance Company  advocates  seamless  brass  tubing,  iron  pipe  size  and 
brass  ground-joint  unions.)  The  fittings  for  lower  or  water  end  should 
be  crosses  or  tees  with  brass  plugs  for  ready  removal  in  cleaning  and 
inspection. 

Valves  should  not  be  put  on  the  pipes  between  the  boiler  and  col- 
umns, as  these  are  a  possible  source  of  danger.  By  making  the  "blow- 
off"  pipe  to  column  %"  or  1"  there  is  no  need  of  any  valves  at  this 
dangerous  point. 

The  lower  gage  cock  and  bottom  of  water  glass  should  be  placed  on 
a  line  3"  above  top  row  of  tubes;  that  is,  water  should  appear  in  the 
glass  when  tubes  are  covered  to  a  depth  of  3". 


Q.  100.  (1898-9.)  Give  size  and  specify  the  number  and  style  of 
safety  valves  necessary  for  the  boilers  referred  to  in  Q.  93 — pop  or  lever 
pattern? 

Ans.  100.  Good  'pop"  safety  valves:  of  standard  make  are  preferred; 
each  boiler  to  have  its  own  valve;  every  valve  to  be  in  direct  communi- 
cation with  the  boiler  which  it  serves  and  placed  with  no  intervening 
means  of  shutting  off  same. 

'ihe  grate  area  to  be  provided  for  either  of  the  boilers  described 
in  Ans.  93  will  hardly  be  less  than  30,  or  more  than  38,  square  feet. 
Figuring  on  the  maximum  grate  area  and  allowing  one  square  inch  of 
valve  area  for  uiree  square  feet  of  grate  we  should  provide  a  valve 
having  38-^3  =  122/3  square  inches,  which  is  nominally  the  area 
corresponding  to  a  4" -diameter. 

(Ed.  Com.)— The  above  is  the  U.  S.  government  allowance  and  is 
only  one  of  the  many  rules  on  this  subject.  There  are  other  modes, 
more  logical,  perhaps;  see  Kent's  Pocket  Book,  1898,  page  723. 

While  it  is  not  a  common  practice  to  provide  two  smaller  valves 
instead  of  one  larger  one,  yet  there  seems  much  in  this  idea  that  is1 
commendable,  especially  when  the  valves  are  set  to  blow  off  at  slightly 
different  pressures. 


Q.  101.     (1898-9.)    What  are  fusible  plugs  made  of  and  where  should 
same  be  placed  in  horizontal  tubular  boilers? 

Ans.  101     Fusible  plugs  are  usually  %  or  1"  pipe  size  with  a  taper 
5  through  them  which  is  filled  with  Banca  tin.    The  tin  is  calculated 

42 


to  fuse  when  not  covered  with  water,  but  certainty  of  action  depends 
on  proper  exposure;  that  is,  scale  should  not  be  allowed  to  accumulate 
on  the  face  of  the  plug. 

The  plugs  should  be  placed  in  the  back  head,  in  rear  combustion 
chamber,  well  exposed  at  highest  point  on  the  fire  line,  which  should 
be  slightly  above  top  row  of  tubes. 


Q.  106.  (1898-9.)  Give  size  of  "stack"  adapted  for  200  HP,  i.  e., 
for  boilers  proposed  in  Q.  93,  height  of  chimney  assumed  at  100  ft. 

Ans.  106.  For  200  HP  boilers  referred  to  in  Q.  No.  93,  the  diameter 
for  chimney  corresponding  to  a  height  of  100  ft.  should  be  38  inches. 
This  assumes  the  consumption  of  5  Ibs.  good  coal  per  HP  per  hour  and 
therefore  provides  for  overload  and  other  contingencies  as  to  the  quality 
of  the  fuel.  (Kent's  Formula.) 


Q.  107.  (1898-9.)  What  is  claimed  for  the  modern  steel  chimney, 
as  compared  with  well-constructed  brick  chimneys? 

Ans.  107.  The  advantages  claimed  for  the  modern  steel  chimney  over 
similar  brick  structures  are:  Greater  strength;  less  space  required; 
foundations  smaller;  the  cost  is  less,  and  furthermore,  the  draft  is  not 
impaired  by  infiltration  of  air.  Steel  chimneys  are  smaller,  lighter 
and  offer  less  area  to  wind  pressure  and  are  also  better  adapted  to 
stand  the  strains,  due  to  unequal  temperatures,  etc. 


Q.  108.  (1898-9.)  Explain  the  distinction  between  "forced"  and 
"induced"  draught;  state  briefly  the  advantages  of  both  systems  when 
acquired  "mechanically,"  i.  e.,  by  fan  blowers. 

Ans.  108.  Under  "forced"  draft  the  air  is  forced  through  the  fire, 
the  pressure  above  the  atmosphere  being  maintained  either  in  the 
closed  ash-pit  or  in  the  closed  fire-room.  The  latter  arrangement  is  only 
practical  in  marine  service. 

Under  the  "induced"  or  suction  method  the  products  of  combustion 
are  exhausted  from  the  furnace,  a  partial  vacuum  is  produced  therein 
and  air  thereby  caused  to  flow  through  the  fuel. 

Both  systems  of  "mechanical  draft"  operate  to  produce  a  pressure 
difference  between  the  ash-pit  and  the  combustion  chamber,  and  other 
things  being  equal,  there  is  no  conclusive  evidence  that  one  method 
is  superior  to  the  other;  the  one  to  be  adopted  must  depend  upon  the 
conditions  and  the  advantages  as  compared  with  "chimney  draft"  are 
common  to  either. 

In  operation,  the  primary  advantage  of  a  mechanical  system  lies  in 
its  intensity  and  its  controllability.  Intense  draft  such  as  may  be  read- 
ily produced  by  means  of  a  fan  makes  possible  the  burning  of  the 
cheapest  fuel,  which  is  always  mostly  in  fine  particles  and  packs  closely 
on  the  grate.  It  is  also  conducive  to  higher  rates  of  combustion  and 
the  carrying  of  deeper  fires.  A  deep  fire  gives  better  opportunity  for 
contact  between  the  air  and  the  fuel  so  that  the  air  supply  can  be 
reduced  and  the  efficiency  of  the  plant  increased. 

Ample  draft  for  the  prevention  of  smoke  can  be  best  produced  by 
mechanical  means,  which  at  the  same  time  prevents  the  formation  of 
carbonic  oxide. 

With  a  chimney  the  temperature  of  the  gases  must  be  high  to  insure 
sufficient  draft,  whereas  with  the  fan  the  gases  may  be  cooled  to  the 
lowest  possible  degree  and  the  heat  transferred  to  the  feed  water  or  the 
air  supply,  thereby  greatly  reducing  the  loss  usually  resulting  from  this 
cause.  It  is  claimed  also  that  the  exhaust  steam  from  the  fan  engines 
may  be  utilized  and  the  expense  of  producing  draft  by  the  heat  savings 
noted,  is  reduced  to  practically  nothing. 

43 


Mechanical  draft  can  be  automatically  regulated  to  instantly  change 
both  the  intensity  of  the  draft  and  the  volume  of  the  air.  It  is  thus 
entirely  independent  of  the  conditions  of  weather  and  is  capable  of 
responding  to  sudden  demands.  By  its  use  a  given  boiler  plant  may  be 
instantly  and  greatly  increased  in  capacity  to  meet  such  demands  as 
are  common  in  electric  railway  work.  Reserve  capacity  can  thus  be 
stored  in  light  and  comparatively  cheap  fans  rather  than  in  ponderous 
and  expensive  boiiers.  When  the  plant  is  properly  designed  less 
boilers  are  required  than  with  the  ordinary  chimney,  for  a  higher  com- 
bustion rate  and  greater  evaporative  capacity  can  be  secured. 


Q.  109.  (1898-9.)  What  difficulties  attending  "hand  firing"  are  over- 
come by  mechanical  stokers? 

Ans.  109.  Mechanical  "stokers"  feed  the  fuel  with  constant  regular- 
ity; they  obviate  the  frequent  opening  of  fire  doors  with  the  usual 
inrush  of  cold  air;  the  fires  are  practically  self-cleaning,  and  by  properly 
arranged  coal  and  ash  handling  machinery  the  cost  of  steam  production 
may  be  much  reduced. 

Stokers  have  to  be  judiciously  selected,  maintained  and  properly 
run  or  they  may  be  a  source  of  annoyance  and  expense. 


Q.  110.  (1898-9.)  Name  what  considerations  should  govern  the  dis- 
tance between  grate  and  boiler. 

Ans.  110.  In  furnace  construction  the  distance  from  grate  to  boiler 
shell  should  be  governed  by  the  kind  of  coal  to  be  burned.  Anthracite 
coal  requires  but  20"  to  24  for  the  best  results;  for  semi-bituminous 
coal  the  distance  should  be  between  26"  to  30"  and  more  in  proportion 
for  rich  bituminous  coals. 


Q.  111.  (1898-9.)  What  portions  of  a  horizontal  tubular  boiler 
should  be  exposed  to  the  fire  and  heated  gases? 

Ans.  111.  Only  such  parts  of  a  horizontal  tubular  boiler  as  are  pro- 
tected by  water  should  be  exposed  to  the  fire  or  gases  of  combustion. 
This  ordinarily  includes  the  lower  3/5  of  the  shell  and  such  surfaces  as 
are  concerned  therewith.  The  return  of  gases  over  top  of  shell  does  not 
meet  with  general  favor. 


Q.  112.  (1898-9.)  Explain  the  meaning  of  "ultimate  tensile  strength  ' 
and  "elastic  limit"  and  state  what  proportions  the  latter  should  bear  to 
the  former,  in  good  steel  boiler  plate. 

Ans.  112.  "Ultimate  tensile  strength"  is  the  maximum  stress  that  a 
piece  will  endure  before  being  torn  asunder.  This  strength  depends 
somewhat  upon  the  mode  of  testing;  the  more  rapidly  the  testing  pro- 
ceeds, the  higher  will  be  the  apparent  strength. 

Elastic  limit  defines  the  limiting  strength,  i.  e.,  a  load  greater  than 
expressed  by  the  elastic  limit  will  produce  a  permanent  elongation  or 
"set"  after  the  load  is  removed. 

The  elastic  limit  for  the  best  grades  of  boiler  steel  should  be  50%  of 
e  tensile  strength;  the  "elongation"  varies  with  the  length  and  thick- 
ness of  the  specimen,  but  is  usually  fixed  at  25%  for  plate  %"  thickness 
and  thereabouts. 


(1899"1900-)     Wha^  is  the  effect  of  increasing  the  height  of  a 
How  is  the  difference  in  pressure  measured? 

'  the  C0mmon  draft  of  a  chimney  as  measured  in  inches  of 


44 


What  is  the  effect  of  an  increase  in  the  difference  of  pressure  or  in- 
tensity of  draft? 

What  relation  would  the  increase  of  pressure  or  intensity  of  draft 
have  upon  the  velocity  of  the  gases  which  flow  through  the  chimney? 

Ans.  77.  Increasing  the  height  of  a  chimney  increases  the  intensity 
of  the  draft,  because  it  makes  a  greater  difference  per  unit  of  base 
between  the  weight  of  the  inside  column  of  gases  and  a  column  of 
equal  height  of  the  outside  air. 

The  difference  in  pressure  is  measured  by  the  height  of  the  column 
of  water  it  will  support;  for  such  a  test,  a  U-shaped  tube  has  one  leg 
connected  with  the  inside  of  the  chimney  and  the  other  open  to  the 
air;  the  greater  pressure  of  the  atmosphere  pushes  the  water  toward 
the  chimney  until  the  difference  of  the  heights  of  the  two  columns  in 
the  glass  tube  is  sufficient  to  balance  the  difference  in  pressure  between 
the  flue  and  atmosphere.  The  pressure  is  measured  by  taking  the  dif- 
ference between  the  heights  of  the  two  columns. 

The  common  draft  of  a  chimney  is  about  Vz  in.  for  ordinary  heights 
and  temperatures. 

The  effect  of  an  increase  in  pressure,  or  intensity  of  draft,  is  to 
increase  the  velocity  with  which  the  gases  flow  through  the  chimney. 

The  velocity  increases  as  the  square  root  of  the  pressure;  in  order 
to  double  the  velocity,  the  pressure  would  have  to  be  increased  4  times; 
to  get  3  times  the  velocity,  the  pressure  would  have  to  be  increased 
9  times. 


Q.  78.  (1899-1900.)  What  relation  has  the  height  of  a  chimney  to 
its  capacity? 

The  area  of  a  chimney  to  its  capacity? 

How  does  the  area  increase? 

Give  general  statement  for  the  capacity  of  chimney.  What  effect 
has  temperature  upon  the  capacity? 

Ans.  78.  The  capacity  of  a  chimney  varies  as  the  square  root  of 
the  height. 

In  a  chimney  200  ft.  in  height,  the  velocity  will  be,  theoretically, 
twice  as  much  as  a  chimney  50  ft.  high,  so  that  twice  as  much  gas 
would  be  discharged  in  a  given  time. 

The  capacity  varies  directly  as  the  area — that  is,  a  chimney  of 
twice  the  cross-section  will  discharge  twice  the  amount  of  gas  in  a 
given  time. 

The  area  increases  as  the  square  of  the  diameter.  The  capacity 
varies  as  the  square  root  of  the  height  and  as  the  square  of  the  diam- 
eter. 

The  effect  of  temperature  on  the  capacity  is  that  the  greater  the 
inside  temperature,  the  greater  the  difference  in  pressure  and  velocity 
of  the  gases;  but,  as  the  density  of  the  gases  decreases  with  the  tem- 
perature, there  is  a  point  where  as  much  is  lost  in  weight  of  the  gas 
passed,  by  the  lightness  of  the  gas,  as  is  gained  by  the  increased  ve- 
locity. 


Q.  79.  (1899-1900.)  How  would  you  find  the  area  required  for  a 
chimney  of  a  given  height  and  the  number  of  sq.  ft.  of  grate  surface 
connected?  How  could  the  area  for  a  given  H.P.  be  determined? 

Ans.  79.  Multiply  the  number  of  sq.  ft.  of  grate  surface  by  120,  and 
divide  the  product  by  the  square  root  of  the  height  and  the  quotient 
will  be  the  required  cross-section  in  square  inches. 

Divide  the  H.P.  by  3  1-3  times  the  square  root  of  the  height,  and 
the  quotient  will  be  the  required  effective  area  in  square  feet. 

To  the  diameter,  or  length  of  the  required  side  to  give  this  area, 
add  4  in.,  to  compensate  for  friction. 

45 


O    80    (1899-1900.)     What  would  be  the  area  of  a  chimney  for  55 
sq.  ft.  of  grate  surface,  height  being  125  ft.,  allowing  for  burning  5  Ibs. 

°f  No^e-Th^'allowancTof  5  Ibs.  of  coal  per  H.P.  per  hour  is  usually 
made  for  difference  in  condition  in  the  atmosphere;  also  for  an  in- 
crease in  demand  for  extra  power,  and  is  the  usual  base  for  formulat- 
ing. 

Ans.  80. 

125  ft.  =  height  of  the  chimney. 

55  sq.  ft.  =  grate  surface. 
55  X  120 

—  =  595  sq.  in.  =  area  of  cross-section. 

1/125 

/    595 
Diameter  =  ^  / =    27.4  in. 


;   595 
"\  .7854 


Adding  4"  for  friction,  diameter  27.444  =  31.4  in. 

Should  a  square  chimney  be  called  for,  then  the  square  root  of  the 
area  (595  in.)  would  equal  the  length  of  the  required  side,  adding  the 
4  in.  as  before. 

Should  a  parallelogram  be  required  in  the  cross-section  of  the  chim- 
ney, having  the  area  (595  sq.  in.)  given,  divide  the  area  by  the  estab- 
lished length  of  side,  for  the  side  adjacent,  adding  the  4  in.,  as  before, 
to  both  sides. 


Q.  91.  (1899-1900.)  What  pressure  (draft)  will  a  chimney  135  ft. 
high  produce?  Outside  temperature  62°  F.,  temperature  of  flue  gases 
580°  F. 

Ans.  91.  The  province  of  a  chimney  is  to  serve  a  double  purpose  of 
creating  a  movement  of  air,  or  as  commonly  called,  creating  a  draft,  and 
conducting  away  the  products  of  combustion  or  obnoxious  gases. 

The  draft  depends  upon  the  gases  having  a  higher  temperature,  con- 
sequently lighter  than  an  equal  column  of  outside  air.  The  air  flowing 
or  moving  from  the  place  of  the  higher  pressure  to  that  of  the  lower. 

The  intensity  of  the  draft  depends  upon  the  height  of  the  stack  and 
the  difference  in  temperatures  of  the  outside  air  and  the  hot  gases  in 
the  stack,  called  the  pressure  difference. 

As  an  illustration:  Take  a  stack  100  ft.  high,  outside  air  at  a  temp, 
of  62°  F.,  the  gases  at  a  temp,  of  570°  F. 

A  pound  of  burned  gas  at  62°  F.  has  a  volume  of  about  12.5  cu.  ft, 
while  the  same  gas  at  570°  F.  will  have  a  volume  of  22.27  cu.  ft. 
12.5  (570° +  460°) 

—  =  22.27  cu.  ft. 
62°  +  460° 

The  460°  representing  absolute  temperature  (some  works  use  461°  as 
absolute  temperature). 

A  column  of  gas  100  ft.  high  with  a  base  of  one  square  foot  equals 
a  pressure  of  100  -=-  22.27  =  4.49  pounds  per  square  foot. 

A  pound  of  air  at  62°  F.  has  a  volume  of  13.14  cu.  ft.,  and  a  column 
100  ft.  high  and  one  foot  square  in  section  will  weigh    100  -~  13.14  = 
7.61  pounds.     The  pressure  difference  equals  7.61  —  4.49  =  3.12  pounds 
per  sq.  foot.    Equal  in  oz.  pressure  to 
3.12  X  16 

144 

1  oz.  =  1.7295  in.  of  water. 

.35  oz.  X  1.7295  =  .605  the  draft  pressure  in  inches  of  water. 
It  will  be  seen  that  the  higher  the  stack  the  lower  will  be  the  tem- 
erature  of  the  gases  in  order  to  obtain  the  same  pressure  difference, 
or  the  same  draft. 

46 


The  pressure  (draft)   of  a  chimney  135  ft.  high;  outside  temp.  62C 
F. ;  gases  temp.  580°  F.,  may  be  found  by  the  following  formula: 

7.6           7.9 
P  =  H( )=  .945  inches  of  water. 


(7.6           7.9  \ 
1  =  .945  inches  of 
T            Ta/ 
Substituting — 

/  7.6    7.9  \ 

P  _  135! \-  135  (.0146  —  .0076  =  135  X  .0070  =  .945  inches  of 

V  522         1040  / 

water,  the  pressure  or  draft  of  the  chimney. 
P  =  draft  in  inches  of  water. 
H  =  height  of  chimney. 

T  =  absolute  temp,  outside  air    (460°  +  62°)  =522°   F. 
Ta  =  absolute  temp,  chimney  gases    (460  +  580) )  =1040°   F. 
7.6  and  7.9  constants  obtained  by  experiment. 


Q.  92.  (1899-1900.)  A  chimney  120  ft.  high,  temperature  flue  gases 
660°  F.  It  is  desirable  to  add  an  economizer,  which  will  reduce  the 
temperature  of  the  gases  to  390°  F.  To  what  height  should  the  chim- 
ney be  raised  in  order  that  the  draft  will  remain  the  same  as  at  first? 
Outside  temperature  in  both  cases  60°  F. 
Ans.  92.  H  =  height  chimney. 

H'  =  height  chimney  after  economizer  is  installed. 

T  =  absolute  temp,  outside  air,  460°  -+-  60°  =  520°  F. 

Ta  =  absolute  temp,   gases,   460°  +  660°  =  1120°    F. 

Tb  =  absolute  temp,   gases   after   economizer   is  installed,    460°  + 
390°  =  850°    F. 

P  =  pressure  in  inches  of  water. 

7.6  and  7.9  are  constant  numbers  found  in  part,  by  experiment  and 
mathematical  process. 

(7.6          7.9  \ 
I  =  P  before  economizer  was  installed. 
T             Ta/ 
(7.6          7.9  \ 
I  =  P  after  economizer  is  installed. 
T            Tb/ 


H'  ==  7 . 6  7.9  the  height  of  chimney  after  economizer  is  installed. 

T          Tb 
Substituting — 

/  7.6  7.9  \ 

P  =  120  (  -  )  =  120  (.0146  —  .0071)  =  120  X  .0075  =  .90,  the 

V  520         1120/ 
pressure  in  inches  of  water. 
.90 

.90  .90 

H'=  7.6         7.9  =—  —  =  —  =  169.81   ft. 

.0146  — .0093       .0053 
520         850 

the  height  of  the  chimney  after  the  economizer  has  been  installed. 

By  the  use  of  another  formula  we  obtain  an  answer  of  168.35  ft.; 
second,  168.88  ft;  third,  170.52. 

A  slight  variation  in  the  constant  numbers  probably  in  disuse  of 
decimals  make  the  slight  variation  in  the  different  answers. 

47 


Q  94  (1899-1900.)  A  boiler  evaporates  12,468  Ibs.  of  water  per 
hour'  (feed-water  154°  F.)  to  a  steam  pressure  of  145  Ibs.  (absolute) 
per  square  inch,  find  the  equivalent  evaporation? 

Ans  94  The  total  heat  of  steam,  divided  by  the  latent  heat  of 
evaporation  of  water,  at  212°  F.,  gives  a  multiplier,  by  which  the 
weight  of  water  actually  evaporated  is  to  be  multiplied(by,  to  reduce  it 
to  the  equivalent  evaporation  of,  "from  and  at  212°  F." 

This  total  heat  of  steam  is  modified  somewhat  by  circumstances. 

From  the  total  heat  of  steam,  subtract  the  heat  units  in  the  feed 
water,  and  divide  the  remainder  by  the  latent  heat  of  evaporation 
at  212°  F. 

Solution — 

1190.4  B.  T.  U.  equals  the  number  of  heat  units  in  steam  at  145 
pounds  pressure  absolute. 

122.33  B.  T.  U.  equals  the  number  of  heat  units  in  water  at  154°  F. 

1190.4  — 122.33  =  1068.07,  the  total  heat. 

1068.07  -4-  966  =  1.1056,  the  factor  of  evaporation. 

12468  Ibs.  steam  X  1.1056  =  13784.62  Ibs.,  "from  and  at  212°  F.,"  the 
equivalent  evaporation. 


Q.  5.  (1900-1901.)  Assume  you  are  operating  a  plant  furnishing 
steam  for  purposes  other  than  supplying  steam  for  your  engine,  how 
would  you  arrange  your  coal  report  so  as  to  charge  each  department 
with  the  proper  amount  of  fuel  consumed? 

Ans.  5.  To  answer  this  question  other  than  as  an  approximation  is 
beyond  the  limit  of  a  work  of  this  character.  There  are  plants  of  this 
kind  where  the  operator  generally  takes  such  methods  as  he  may  be 
possessed  of  for  forming  approximate  ideas  relative  to  the  coal  and 
water  consumption. 

The  indicated  power  is,  of  course,  first  taken  care  of  and  the 
balance  is  subdivided  for  different  purposes.  The  first  thing  requisite 
is  good  judgment,  and  a  careful  scrutiny  of  the  per  cent  loss  not 
returned  in  condensation,  etc.  By  measuring  the  returning  condensa- 
tion water  and  steam  for  each  system  (by  meter  or  otherwise). 

If  the  returns  are  returned  to  the  boiler,  then  the  difference  be- 
tween the  heat  units  in  the  steam  and  the  heat  units  in  the  returns 
will  equal  the  heat  units  used;  these  multiplied  by  the  number  of 
pounds  of  returned  condensation  equal  the  total  number  of  heat  units 
used;  divide  this  product  by  the  effective  heat  units  in  one  pound  of 
coal  consumed  and  the  quotient  will  equal  the  number  of  pounds  of 
coal  used. 

Or,  the  number  of  pounds  of  returned  condensation  divided  by  the 
number  of  pounds  of  water  evaporated  will  also  give  the  number  of 
pounds  of  coal  used,  providing  proper  allowance  is  made  for  the  heat 
that  is  returned  to  the  boiler. 


Q.  6.  (1900-1901.)  Give  opinion  relative  to  the  several  advantages 
and  disadvantages  of  the  two  general  types  of  boilers,  water  tube  and 
fire  tube. 

Ans.  6.  The  water  tube  boiler,  while  not  rated  at  as  high  efficiency 
as  the  return  tubular  boiler,  has  an  element  of  safety  superior  to  the 
return  tubular  boiler. 

It  is  the  tendency  at  the  present  day  with  the  multi-cylinder  engines 

to  carry  extreme  high   pressures,   especially   in   the   marine'  service. 

ucn  boilers  must  be  carefully  designed  for  strength.     It  is  likewise 

necessary  to  reduce  its  weight  and  size  to  the  lowest  possible  limit, 

e  best  of  material  should  be  used.    For  carrying  high  pressures, 

is  clear  to  see,  for  the  sake  of  safety,  the  Scotch  type  or  any  tubular 

Her  must  be  made  extremely  heavy  and  bulky,  and  much 

48 


attention  is  now  being  paid  to  the  devising  of  a  new  type,  which, 
while  retaining  the  good  features  of  the  Scotch  type,  will  be  lighter, 
smaller  and  cheaper  for  the  same  power. 

Some  of  the  advantages  claimed  for  water-tube  boilers  are  as 
follows: 

1. — That  the  portions  of  the  boiler  which  contain  the  water  are  so 
small  in  diameter  that  the  material  used  in  the  construction  can  be 
made  comparatively  light  without  impairing  the  strength.  Conse- 
quently the  heat  is  transmitted  to  the  water  more  readily  and  the 
danger  of  burning  the  iron  where  it  is  exposed  to  the  fire  is  greatly 
diminished, 

2. — There  are  no  riveted  joints,  and  consequently  no  double  thick- 
ness of  metal  exposed  to  the  fire. 

3. — That  the  draught  area,  being  much  larger  than  in  the  fire-tube 
boilers,  gives  ample  time  for  the  absorption  of  the  heat  of  the  gases 
before  their  exit  to  the  chimney. 

4. — That  the  gases  being  thoroughly  mingled  in  their  passage  be- 
tween the  staggered  tubes,  the  combustion  is  more  complete. 

5.— That  the  gases  impinging  against  the  heating  surface  perpen- 
dicularly instead  of  gliding  along  the  same  longitudinally,  the  absorp- 
tion of  heat  is  more  thorough.  It  is  claimed  that  experiments  have 
proven  that  a  gain  of  30  per  cent  in  the  efficiency  of  the  heating  sur- 
face is  effected. 

6. — That  the  circulation  of  water  is  rapid  and  all  in  the  same  direc- 
tion, there  being  no  conflicting  currents.  As  a  result  the  temperature 
of  the  boiler  in  all  its  parts  is  practically  the  same  and  the  tendency 
to  deposit  scale  is  materially  lessened. 

7. — That  the  water  being  divided  into  many  small  streams  in  thin 
envelopes,  steam  may  be  raised  rapidly. 

8. — That  the  large  area  of  disengaging  surface  in  the  drums, 
together  with  the  fact  that  the  steam  is  delivered  at  one  end  and 
taken  out  at  the  other  end,  insures  dry  steam  without  the  aid  of  any 
superheating  device. 

9. — That  the  water  level  may  readily  be  kept  steady. 

10.— That  the  whole  structure  is  so  flexible  that  the  parts  may 
expand  and  contract  without  producing  strains. 

11. — That  the  division  of  water  into  small  masses  avoids  destructive 
explosions. 

12. — That  the  space  occupied  by  this  type  of  boiler  for  a  given 
power  is  much  less  than  in  fire-tube  boilers. 

13. — That  by  a  suitable  arrangement  of  hand  and  man  holes  every 
part  of  the  boiler  is  accessible  for  cleaning  and  repairs. 

14. — That  the  loss  of  effect  from  dust  collecting  on  the  top  of  the 
tubes  is  far  less  than  in  fire-tube  boilers,  where  it  collects  on  the 
interior  surface.  In  the  latter  case  there  is  no  limit  to  the  amount 
of  dust  which  may  collect,  while  in  the  former  only  a  limited  amount 
is  retained. 

15. — That  since  no  part  of  the  boiler  above  the  water  level  is  ex- 
posed to  the  fire,  and  because  of  the  absence  of  deteriorating  strains 
and  thick  plates  and  joints  in  the  fire,  it  is  much  more  durable. 

16. — That,  being  made  in  sections,  it  is  less  cumbrous  and  much 
more  easily  transported  and  erected. 

17. — That  a  new  tube  may  be  easily  inserted  or  most  any  other 
repair  be  made  by  an  ordinary  mechanic  with  ordinary  tools. 


Q.  7.  (1900-1901.)     Give  rules  for  selecting  material  for  cylindrical 
shells,  also  shell  plate  formula.    Also  rules  for  flat  plates. 

Give   furnace   formula. 

Give  rules  for  selecting  stays,  also  rules  for  allowable  safe  load  on 
stays. 

Material  for  cylindrical  shells  subject  to  internal  pressure. 
49 


Ans.  7.  Board  of  Trade  Rules.— Tensile  strength  between  27  and  32 
tons.  In  the  normal  condition  the  elongation  is  not  less  than  18  per  cent 
in  10  inches,  but  should  be  about  25  per  cent;  if  annealed  not  less  than 
20  per  cent.  Strips  2  inches  wide  should  stand  bending  until  the  sides 
are  parallel  at  a  distance  from  each  other  of  not  more  than  three  times 
the  thickness  of  the  material. 

Lloyd's. — Tensile  strength  between  the  limits  of  26  and  30  tons 
per  sq.  in.  Elongation  not  less  than  20  per  cent  in  8  inches.  Test 
strips  heated  to  a  low  cherry-red  and  plunged  into  water  at  82°  F. 
Must  stand  bending  to  a  curve  the  inner  radius  of  which  is  not 
greater  than  1%  times  the  plates'  thickness. 

U.  S.  Statutes. — Plates  of  %-inch  thickness  and  under  shall  show 
a  contraction  of  area  of  not  less  than  50  per  cent;  when  over  %  inch 
and  up  to  %  inch  not  less  than  45  per  cent;  when  over  %  inch  not  less 
than  40  per  cent. 

From  a  paper  on  boiler  construction  by  Nelson  Foley,  we  find  the 
following  comments:  "The  Board  of  Trade  rules  seem  to  indicate  a 
steel  of  too  high  a  tensile  strength,  when  a  lower  and  more  ductile 
one  can  be  got;  the  lower  tensile  strength  limit  should  be  reduced, 
and  the  bending  test  might,  with  advantage,  be  made  after  tempering, 
and  made  to  a  smaller  radius.  Lloyd's  rule  for  quality  seems  more 
satisfactory,  but  the  temper  test  is  not  severe.  The  U.  S.  statutes  are 
not  sufficiently  stringent  to  insure  an  entirely  satisfactory  material." 

Mr.  Foley  suggests  a  material  which  would  meet  the  following: 
25  tons  the  lowest  limit  in  tension;  25  per  cent  in  8  inches  minimum 
elongation;  radius  for  bending  after  tempering  equals  the  plate's 
thickness. 

Shell  plate  formula: 

Board  of  Trade: 

TxBxtx2  2  (TBt) 

P  = orP  = . 

DxFxlOO  100  (DF) 

D  =  diameter  of  boiler  in  inches; 

P  =  working  pressure  in  Ibs.  per  sq.  in.; 

t  =  thickness  in  inches; 

B  =  percentage  of  strength  of  joint  compared  to  solid  plate; 

T  =  tensile  strength  allowed  for  material  in  Ibs.  for  sq.  in.; 

F  =  factor  of  safety;    being  4.5,  with  certain  additions  depending 
upon  method  of  construction. 
C  (t  — 2)  B 

Lloyd's:    P  =  - 

D 

t=  thickness  of  plate  in  sixteenths. 

B  and  D  =  same  as  in  previous  formula. 

C  =  constant  depending  on  the  kind  of  joint. 

When  longitudinal  seams  have  double  butt  straps,  C  —  20. 

When  longitudinal  seams  have  double  butt  straps  of  unequal  width, 
only  covering  on  one  side  the  reduced  section  of  plate  at  the  outer 
line  of  rivets,  C  =  19.5. 

When  longitudinal  seams  are  lap-jointed    C  =  18  5 
2tT 

U.  S.  statutes:    P  =  —    —  for  single  riveting;  add  20  per  cent  for 
6D 

S6  th6  Sam6  notation  as  in  Board  of  Trade  rules  or 
criticises  the  u-  S.  statutes  as  follows:  "The  rule  ignores 
tf  °e?n  that  *  distlnguishes  between  single  and  double, 
the  latter  20  per  cent  advantage;   the  circumferential  riveting 
of  »   X™     S^mS  1S  alt°sether  ignored.     The  rule  takes  no  account 
^    •  +LP   °r  meth°d  adopted   of  construction   of  joints.     The 
-sixth  simply  covers  the  actual   nominal   factor   of  safety, 

5° 


as  well  as  the  loss  of  strength  at  the  joint,  no  matter  what  the  per- 
centage;  we  may  therefore  dismiss  it  as  unsatisfactory-" 
Plat  Plates — 

The  Board  of  Trade  rules  for  flat  surfaces,  being  based  on  actual 
experiment,  are  especially  worthy  of  respect;  sound  judgment  appears 
also  to  have  been  used  in  framing  them,  and  will  be  the  only  ones 
given  in  this  work. 
Board  of  Trade: 

C  (t  +  1)  2 

P  = 

S  — 6 

p  =  working  pressure  in  Ibs.  per  sq.  in. 
S  =  surface  supported  in  sq.  in. 
t  =  thickness  in  sixteenths  of  an  in. 
C  =  constant. 

C  =  125  for  plates  not  exposed  to  heat  or  flame,  the  stays  fitted  with 
nuts  and  washers,  the  latter  three  times  the  diameter  of  the  stay  and 
two-thirds  the  thickness  of  the  plate. 

C  =  187.5  for  the  same  conditions,  but  the  washers  two-thirds  the 
pitch  of  the  stays,  and  thickness  the  same  as  the  plate. 

C  =  200  for  the  same  conditions,  but  doubling  plates  in  place  of 
washers,  the  width  of  which  is  two-thirds  the  pitch  and  the  thickness 
the  same  as  the  plate. 

0  =  112.5  for  the  same  condition,  but  the  stays  with  nuts  only. 
C  =  75  when  exposed  to  impact  heat  or  flame  and  steam  in  contact 
with  the  plates,  and  the  stays  fitted  with  nuts  and  washers  three  times 
the  diameter  of  the  stay  and  two-thirds  the  plate's  thickness. 

0  =  67.5  when  the  same  condition  exists,  but  the  stays  fitted  with 
nuts  only. 

0  =  100  when  exposed  to  heat  or  flame  and  water  in  contact  with 
the  plates,  and  stays  screwed  into  the  plates  and  fitted  with  nuts. 
C  =  66  for  the  same  condition,  but  stays  with  riveted  heads. 
Furnace  Formula:  — 
Board  of  Trade. — Long  Furnaces: 
Ct2 

P  = ,   but  not  where  L   is  shorter  than    (ll.St  —  1),   at 

(L  +  l)  D 

which  length  the  rule  for  short  furnaces  comes  into  play. 
P  =  working  pressure  in  Ibs.  per  sq.  in.; 
t  =  thickness   in   inches; 
D  =  outside  diameter  in  inches; 
L  = length  of  furnace  up  to  10  ft.; 
0  =  constant  as  per  following  table  for  drilled  holes: 
0  =  99000   for   welded    or   butt-jointed    with    single    straps,    double 
riveted; 

C  =  88000   for  butts  with  single  straps   single  riveted; 
0  =  99000  for  butts  with  double  straps  single  riveted. 
Provided,  however,  that  the  pressure  so  found  does  not  exceed  that 
given  by  the  following  formula,  which  applies  also  to  short  furnaces: 
Ct 

P  = —  for  all  patent  furnaces  named; 

D 
Ct  /         12IA 

P  = 15 I  when  with  Adamson  rings. 

3D\       67.5t/ 

0  =  8800   for  plain   furnaces; 

0  =  14000  for  "Fox";  minimum  thickness  5/16  inch,  greatest  5/8 
inch,  plain  part  not  to  exceed  6  inches  in  length; 

0  =  13500  for  "Morrison;"  minimum  thickness  5/16  inch,  greatest 
5/8  inch,  plain  part  not  to  exceed  6  inches  in  length; 


0  =  14000   for   "Purves-Brown;"   limit   of  thickness   7/16   and    5/8 


of  flange  next  nre  1*  inches. 

U.  S.  Statutes.—  Short  Furnaces:    Plain  and  Patent.    When  less  than 
8  ft.  in  length: 
89600t* 

LD 

tc 

P  =  -  when 

C  =  14000  for  Fox  corrugations  where  D  equals  mean  diameter. 
C  =  14000  for  Purves-Brown  where  D  equals  the  diameter  of  the 

6  =  5677  for  plain  flues  over  16  inches  in  diameter  and  less  than 
40  inches,  when  not  over  3  ft.  in  length. 

Material  for  stays:  — 

Board  of  Trade.—  The  tensile  strength  to  lie  within  the  limit  of  27 
and  32  tons  per  square  inch,  and  to  have  an  elongation  of  not  less 
than  20  per  cent  in  10  inches. 

Steel  stays,  which  have  been  welded  or  worked  in  the  fire,  should 
not  be  used. 

Lloyd's.—  26  to  30  tons,  steel,  with  elongation  not  less  than  20  per 
cent  in  8  inches. 

U.  S.  Statutes.—  The  only  condition  is  that  the  reduction  in  area 
must  not  be  less  than  40  per  cent  if  the  test  bar  is  over  %  inch  in 
diameter. 

Safe  loads  on  stays:  — 

Board  of  Trade.  —  9000  Ibs.  per  square  inch  is  allowed  on  the  net 
section,  provided  the  tensile  strength  range  from  27  to  32  tons. 

Steel  stays  are  not  to  be  welded  or  worked  in  the  fire. 

Lloyd's.  —  For  screwed  and  other  stays  not  exceeding  IV2  inches  in 
diameter,  effective,  8000  per  square  inch  is  allowed;  for  stays  above 
1%  inches  9000  Ibs.  No  stays  are  to  be  welded. 

U.  S.  Statutes—  Braces  and  stays  shall  not  be  subjected  to  a  greater 
stress  than  6000  Ibs.  per  square  inch. 


Q.  8.  (1900-01.)  ^ive  rules  for  tube  plates,  finding  thickness  and 
distance  between  tubes;  that  is,  how  much  material  should  properly 
be  left  between  the  tubes? 

Give  rules  for  establishing  the  value  of  material  for  boiler  tubes. 

Would  you  allow  for  the  holding  power  of  boiler  tubes  due  to  fric- 
tion between  the  outer  surface  of  the  tube  and  the  surface  of  the  hole 
in  the  tube  sheet? 

What  do  you  consider  as  the  best  form  of  hole  through  the  tube 


What  would  be  the  relation  in  efficiency  between  a  tube  simply  ex- 
panded in  and  one  expanded  in  and  beaded?  That  is,  as  regards  the 
holding  power  of  tube. 

Ans.  8.    Tube  Plates: 

Board  of  Trade  rule: 

t  (D  —  d)  X  20000 


WD 

D  =  least  horizontal  distance  between  centers  of  tubes  in  inches; 

d  =  inside  diameter  of  ordinary  tubes; 

t  =  thickness  of  tube  plate  in  inches; 

W  =  extreme  width  of  combustion-box  in  inches  from  front  tube- 
plate  to  back  of  fire  box,  or,  distance  between  combustion  box  tube 
plates  when  boiler  is  double-ended  and  the  box  common  to  both  ends; 

P  =  pitch  of  tubes; 

52 


The  thickness  of  tube  plates  is  generally  one-eighth  of  an  inch  in 
excess  of  the  sheet  forming  the  shell  of  the  boiler. 

Material  for  boiler  tubes:  — 

If  of  iron,  the  quality  to  be  such  as  to  give  at  least  22  tons  per 
sq.  in.  as  the  minimum  tensile  strength,  with  an  elongation  of  not  less 
than  15  per  cent  in  8  inches.  If  of  steel,  the  elongation  to  be  not 
less  than  26  per  cent  in  8  inches  for  the  material  before  being  rolled 
into  strips;  and  after  tempering,  the  test  bar  to  stand  completely 
closing  together.  Provided,  the  steel  welds  well,  there  does  not  seem 
to  be  any  objection  in  providing  tensile  limits. 

The  ends  should  be  annealed  after  manufacture  and  stay-tube  ends 
should  be  annealed  before  screwing. 

The  holding  power  of  tubes:  — 

Experiments  made  in  the  Washington  Navy  Yard  show  that,  with 
2%-inch  brass  tubes,  in  no  case  was  the  holding  power  less  than  6000 
Ibs.,  while  the  average  was  upwards  of  20000  Ibs.  It  was  further  shown 
that  with  these  tubes,  nuts  were  superfluous,  quite  as  good  results 
being  obtained  with  tubes  simply  expanded  into  the  tube-sheet  and 
fitted  with  a  ferrule. 

In  five  experiments  with  steel  tubes  2  inches  and  21/4  inches  in 
diameter,  the  first  five  tubes  gave  way  on  an  average  of  23740  Ibs., 
which  would  appear  to  be  about  2/3  the  ultimate  strength  of  the  tubes 
themselves. 

In  all  these  cases  the  hole  through  the  tube-plate  was  parallel  with 
a  sharp  edge  to  it,  and  a  ferrule  was  driven  into  the  tube. 

Another  test  of  five  steel  tubes  made  under  the  same  conditions 
as  the  first  five,  with  the  exception  that  the  ferrule  was  omitted,  the 
tubes  simply  being  expanded  into  the  plate.  The  mean  pull  required 
15270  Ibs.,  or  considerably  less  than  half  the  ultimate  strength  of  the 
tubes. 

The  effect  of  beading  the  tubes,  the  holes  through  the  plate  being 
parallel  and  ferrules  omitted.  The  mean  of  the  first  three,  which  are 
tubes  of  the  same  kind,  gives  26876  Ibs.  for  the  holding  power,  under 
these  conditions,  as  compared  with  23740  Ibs.  for  tubes  fitted  with 
ferrules  only.  This  high  figure  is,  however,  due  mainly  to  an  excep- 
tional case  when  the  holding  power  is  greater  than  the  average  strength 
of  the  tubes  themselves. 

It  is  disadvantageous  to  cone  the  hole  through  the  tube-plate  unless 
the  sharp  edge  is  removed,  as  the  results  are  much  worse  than  those 
obtained  with  a  parallel  hole. 

In  experiments  on  tubes  expanded  into  tapered  holes  and  simply 
beaded  over,  better  results  were  obtained  than  with  ferrules;  in  these 
cases,  however,  the  sharp  edge  of  the  hole  was  rounded  off,  which 
appears  in  general  to  have  a  good  effect. 

Experiments  by  Yarrow  &  Co.: 

In  fifteen  experiments  on  4  and  5-inch  steel  tubes,  the  strain  ranged 
from  20720  to  68040  Ibs.  Beading  the  tubes  does  not  necessarily  give 
increased  resistance,  as  some  of  the  lower  figures  were  obtained  from 
beaded  tubes. 


Q.  9.  (1900-01.)  What  controls  the  diameter  of  rivets,  and  what 
is  the  extreme  limit  of  their  pitch? 

What  should  be  the  thickness  of  double  butt  straps  (each)  and 
what  should  be  the  thickness  of  single  butt  straps? 

What  should  be  the  distance  from  edge  of  plate  to  the  center  of 
rivet  holes? 

What  should  be  the  distance  between  the  rows  of  rivets  when  chain 
riveted  and  when  zig-zag  riveted? 

Ans.  9.     For  Single-Riveted  Plates.— Lap  Joint: 

The  diameter  of  the  hole  should  be  two  and  one-third  (21-3)  times 
the  thickness  of  the  plate,  and  the  pitch  of  the  rivet  two  and  three- 

53 


eighths  times  the  diameter  of  the  hole,  making  the  mean  plate  area  71 
per  cent  of  the  rivet  area. 

For  double-riveted  lap-joints: 

The  ratio  of  diameter  to  thickness  remains  the  same  as  in  single- 
riveted  lap-joints;  while  the  ratio  of  pitch  to  diameter  of  hole  should 
be  3.64  for  30-ton  plates,  and  22  and  24  ton  rivets,  and  3.82  for  28-ton 
plates  with  the  same  rivets. 

The  distance  from  the  edge  of  plate  to  center  of  rivet  hole  should 
be  1%  diameters. 

The  thickness  of  double  butt  straps  should  be  at  least  %  of  the 
thickness  of  the  plate,  and  for  single-butt  straps  the  thickness  should 
be  1%  times  the  thickness  of  the  plate. 

"Kent"  gives  the  following  formula  for  the  pitch: 

Single  riveted  plates  P  =  .571 h  a 

d1 

Double  riveted  plates  r  =  1.142 f-  d 

t 

P  =  pitch  of  the  rivets; 
d==  diameter  of  hole; 
t  =  thickness  of  plate; 

The  co-efficients  .571  and  1.142  agree  closely  with  those  given  in  the 
report  of  the  committee  of  the  Institution  of  M.  B. 
Distance  between  rows  of  rivets  in  chain  riveting  = 
[(diam.  X  4)  +1] 

D  =  2  X  diam.  of  rivet,  or 

2 

Zigzag  = ; 

D=  V  [(pitch  X  11)  +  (diam.  X  4)]  X  (pitch  +  diam.  X  4) 

(pitch  X  6  + diam.  X  4) 
Diagonal  pitch  = 

10 

Note. — The  subject  of  riveting,  in  its  many  forms  of  joints  and  con- 
ditions, is  exhaustless,  and  to  do  proper  justice  to  the  subject  is 
not  within  the  province  of  this  work,  and  to  those  seeking  a  thorough 
knowledge  on  this  subject,  the  standard  authorities  should  be  consulted. 


Q.  10.  (1900-01.)  What  is  your  opinion  of  iron  versus  steel  boiler 
tubes? 

Give  rule  for  finding  allowable. pressure  on  bumped  heads  of  boil- 
ers. 

Ans.  10.    A  good  grade  of  charcoal  iron  makes  the  best  boiler  tube. 

Mild  homogenous  steel  is  used  to  a  great  extent  and  makes  a  very 
good  tube. 

If  wrought  iron  is  used,  it  should  have  a  tensile  strength  of  not 
less  than  45000  pounds  per  sq.  in.,  and  an  elongation  of  15  per  cent 
in  8  inches,  and  after  tempering  the  test  bar  should  stand  completely 
closing  together. 

Experiments  seem  to  indicate  that,  so  far  as  leakage  is  concerned, 
iron  is  preferable  because  it  is  not  subject  to  the  same  degree  of  ex- 
pansion and  contraction  as  steel. 

Bumped  Heads: 

In  the  construction  of  bumped  heads  for  boilers,  in  order  that  the 
head  should  have  the  same  strength  as  the  shell,  the  head  should  be 
bumped;  that  is,  the  spherical  part  of  the  head  should  be  curved  to 
a  radius  equal  to  the  diameter  of  the  boiler.  Should  a  larger  radius  be 
used  the  tendency  would  be  to  weaken  the  boiler 

The  nearer  hemispherical  the  head  is  the  stronger  it  is. 

54 


Rules  for  allowable  pressure: 

Multiply  the  thickness  of  the  plate  by  1/6  of  the  tensile  strength 
and  divide  this  product  by  6/10  of  the  radius,  to  which  the  head  is 
bumped,  which  will  give  the  pressure  per  sq.  in.  allowable. 


Q.  11.  (1900-01.)  What  should  be  the  tensile  strength  (T.  S.)  elon- 
gation and  contraction  of  area  of  the  materials  for  rivets? 

What  shearing  resistance  per  square  inch,  and  what  factor  of  safety 
should  be  used  for  steel  rivets? 

What  difference  should  be  allowed  in  the  calculations  between  rivets 
in  single  and  double  shear? 

Ans.  11.    Rules  Connected  with  Riveting: 

Board  of  Trade. — The  shearing  resistance  of  the  rivet  steel  to  be 
taken  at  23  tons  per  sq.  in.,  5  to  be  used  as  a  factor  of  safety  independ- 
ently of  any  addition  to  this  factor  for  plating.  Rivets  in  double 
shear  to  have  1.75  times  the  single  section  taken  in  the  calculation 
instead  of  2.  The  diameter  must  not  be  less  than  the  thickness  of  the 
plate  and  the  pitch  never  greater  than  8%  inches. 

Lloyd's. — The  shearing  strength  of  rivet  steel  to  be  taken  at  85  per 
cent  of  the  tensile  strength. 

The  tensile  strength  of  rivet  bars  between  26  and  30  tons,  with  an 
elongation  in  10  inches  of  not  less  than  25  per  cent  and  a  contraction 
in  area  not  less  than  50  per  cent. 


Q.  12.  (1900-01.)  Give  rules  for  proportioning  the  areas  of  flues, 
tubes  and  other  gas  passages  for  both  anthracite  and  bituminous  coals. 

For  air  passages  in  grate  bars. 

Ans.  12.    For  anthracite  coal  1/9  to  1/10  of  the  grate  surface. 

For  bituminous  coal  1/6  to  1/7  of  the  grate  surface. 

The  tube  or  flue  area  should  be  of  sufficient  area  so  as  not  to  impede 
the  passage  of  the  gases  and  not  impair  the  efficiency  of  the  boiler. 

If  too  large,  there  is  a  tendency  for  the  gases  to  select  passages  of 
the  least  resistance,  escaping  without  wholly  giving  up  their  heat, 
also  impairing  the  efficiency  of  the  boiler. 

The  grate  bars  are  usually  constructed  with  an  allowance  of  45  to 
55  per  cent  air  space.  Some  forms  of  construction  require  more  air 
space  than  others. 


Q.  13.  (1900-01.)  How  is  the  working  pressure  of  a  boiler  calcu- 
lated from  the  pressure  of  the  usual  hydrostatic  test? 

What  are  the  several  rules  used  in  establishing  the  nominal  factor 
of  safety  in  boiler  construction? 

Ans.  13.  The  hydrostatic  test,  as  applied  to  boilers,  is  usually  1% 
times  the  working  pressure  of  the  boiler. 

Nelson  Foley  proposes  that  the  proof  pressure  should  be  1%  times 
the  working  pressure  plus  one  atmosphere  (15  Ibs.). 

The  rules  for  finding  factor  of  safety  are  somewhat  conflicting.  The 
factor  of  safety  equals  the  bursting  pressure  of  the  boiler  as  figured 
by  the  rules  before  given,  divided  by  the  working  pressure  for  which 
the  boiler  is  built. 

The  factor  is  usually  considered  in  connection  with  the  tensile 
strength  of  the  material,  the  character  of  the  joint  and  the  workman- 
ship, and  range  from  3%  per  cent  upward,  according  to  conditions. 


Q.  14.  (1900-01.)  The  term  "horse-power"  as  applied  to  steam  boil- 
ers, means  the  capacity  of  a  boiler  to  evaporate  30  pounds  of  water 
from  and  at  a  temperature  of  100°  F.  into  steam  of  84.7  pounds  abso- 
lute pressure  per  square  inch. 

55 


What  should  be  the  average  proportion  for  maximum  economy, 
hand  firing,  good  anthracite  coal,  of — 

Heating  surface  per  horse  power? 

Grate  surface  per  horse-power? 

Ratio  of  heating  to  grate  surface? 

Water  evaporated  from  and  at  212°  F.  per  square  foot  of  heating 
surface  per  hour? 

Combustible  burned  per  horse-power  per  hour? 

Combustible  burned  per  square  foot  of  grate  surface  per  hour? 

Coal  with  16  2/3  %  refuse  in  pounds  per  hour? 

Coal  with  16  2/3  %  refuse  in  pounds  per  hour  per  square  foot  of 
grate  surface? 

Water  evaporated  from  and  at  212°  F.  per  pound  ot  combustible? 

Water  evaporated  from  and  at  212°  F.  per  pound  of  coal,  16  2/3  % 
refuse? 

Ans.  14.  Steam  Boiler  Proportions  (as  per  conditions  of  Question 
15). 

Heating  surface  per  horse-power,  11.5  sq.  ft. 

Grate  surface  per  horse-power,  1/3  sq.  ft. 

Ratio  of  heating  to  grate  surface,  34.5  to  1. 

Water  evp'd  from  and  at  212°  per  sq.  ft.,  H.  S.  per  hour,  3  Ibs. 

Combustible  burned  per  H.  P.,  per  hour,  3  Ibs. 

Coal  with  16  2/3  %  refuse,  Ibs.  per  hour,  3.6  Ibs. 

Per  sq.  ft.  grate  surface,  10.8  Ibs. 

Combustible  burned,  per  sq.  ft.  grate  surface,  9  Ibs. 

Water  evaporated  from  and  at  212°  per  Ib.  comb.,  11.5  Ibs. 

Water  evaporated  from  and  at  212°  per  Ib.  coal,  9.6  Ibs. 


Q.  15.  (190U-01.)  Specify  the  thickness  of  sheets  or  plates,  style  of 
seams,  braces,  size  of  rivets,  pitch,  etc.,  together  with  the  diameter 
and  length  of  shell,  the  diameter  and  number  of  tubes,  for  a  horizontal 
tubular  boiler,  builders  rating  150  horse-power,  maximum  working 
pressure  to  be  120  pounds  per  square  inch,  gauge  pressure. 

Which  is  advisable  to  use,  a  drum  or  nozzle? 

Ans.  15.    Under  conditions  of  the  questions: 

A  boiler  72  inches  in  diameter,  17  feet  long,  132  3-inch  tubes;  heat- 
ing surface  1800  sq.  ft.  The  sheets  %  inch  in  thickness  of  "open  hearth 
fire-box  steel,"  with  tested  tensile  strength  of  60000  Ibs.  per  sq.  in.  of 
section. 

Horizontal  seams— "triple-riveted  double  butt-joint;"  holes  drilled  1 
inch  in  diameter;  rivets  15/16  inch  in  diameter;  pitch  of  rivets  3% 
inches  by  7%  inches.  The  efficiency  of  joints,  86.6  per  cent. 

Through  bracing  from  head  to  head. 

Heating  surfaces: 

6  X  3.1416  X  17 

One-half  shell  = =  160.21  sq.  ft. 

2 

Area  of  heads  =  37.7  sq.  ft. 

Area  of  tube  ends  in  both  heads  equals  6.48  X  2  =  12.96  square  feet. 

Deducted  from  head  area  equals: 

37.68  — 12.96  =  24.72  sq.  ft.  heating  surface  on  heads. 

The  inner  diameter  of  3-inch  tube  equals  2.78  inches. 

Heating  surface  per  ft.  in  length  equals  2.78  X  3.1416  -^  144  =  .7283 
sq.  ft.  per  foot  length. 

Sq.  ft.  heating  surface  in  each  tube  equals  .7283  X  17  =  12.38  sq.  ft. 

Total  heating  surface  in  tubes  equals  12.38  X  132  =  1634.16  sq.  ft 

Total  heating  surface  of  the  boiler: 

Shell,  160.21  sq.  ft;  heads,  24.72  sq.  ft;  tubes,  1634.16  sq.  ft  •  total 
1819.09  sq.  ft.  of  heating  surface. 

The  nozzle  is  used  in  preference  to  the  dome.    The  steam  dome  has 

56 


a  tendency  to  weaken  the  shell  of  the  boiler  and  has  not  proved  of  any 
real  value  in  providing  dry  steam  for  the  engine. 


Q.  16.  (1900-01.)  Give  a  description  of  what  you  consider  a  first- 
class  typical  boiler  setting  for  a  horizontal  tubular  boiler  of  given 
dimensions,  with  total  area  and  height  of  chimney  included  (for 
natural  draft). 

Ans.  16.    Boiler  Setting. — Return  Tubular: 

In  a  boiler  setting,  three  things  are  to  be  obtained: 

First — A  firm  support  for  the  boiler  shell;  this  of  course  includes 
foundations,  walls  and  all  necessary  supports. 

Second — A  properly  proportioned  and  arranged  ash-pit,  furnace  and 
combustion  chamber. 

Third — A  protected  covering  for  the  boiler,  which  shall,  as  far  as 
possible,  prevent  the  loss  of  heat  by  radiation. 

The  Hartford  Boiler  Insurance  Company  describes  the  setting  of 
a  60-inch  horizontal  return-tubular  boiler  as  follows: 

"The  foundation  is  heavy  stone  work  laid  to  the  depth  of  three 
to  four  feet  below  the  surface.  Upon  this  the  brick  wbrk  is  laid. 

"The  side  and  rear  walls  are  double,  with  a  two-inch  air  space 
between  the  inner  and  outer  walls;  there  are  projecting  brick  laid  in 
these  walls,  simply  to  steady  the  brick  work,  but  not  to  interfere  with 
the  expansion  of  the  inner  wall,  caused  by  the  extreme  furnace  heat. 

"The  inside  wall  next  to  the  furnace,  and  hot  gas  passage  is  faced 
with  fire  brick. 

"The  bridge  wall  in  some  instances  is  built  wholly  of  fire  brick;  in 
other  cases  faced. 

"The  boiler  is  supported  by  cast-iron  lugs,  riveted  to  the  shell. 
These  lugs  rest  upon  iron  plates  placed  upon  the  tops  of  the  side 
walls. 

"The  front  lugs  rest  directly  upon  the  plates,  while  the  back  lugs 
rest  upon  rollers  of  one  inch  round  iron  allowing  free  expansion  and 
contraction  of  the  boiler. 

"The  rear  wall  is  24  inches  from  the  rear  head  of  the  boiler,  allow- 
ing ample  space  for  the  gases  to  enter  the  tubes;  above  the  tubes,  how- 
ever, the  wall  is  built  in  to  meet  the  head,  and  forms  a  roof  for  the 
chamber. 

"The  rear  wall  is  provided  with  a  door,  to  remove  the  dirt  and 
soot  that  collects  back  of  the  bridge,  and  also  to  provide  means  for 
inspection. 

"For  anthracite  coal  the  grate  is  placed  24  inches  below  the  shell. 
For  bituminous  28  to  30  inches. 

"The  grate  has  a  fall  of  three  inches  from  front  to  rear,  so  that  the 
fuel  is  thicker  near  the  back  end  of  the  fire;  this  is  believed  to  lead 
to  more  even  combustion,  since  the  air  has  naturally  a  greater  ten- 
dency to  pass  through  the  fire  nearest  the  bridge,  and  upon  meeting  a 
thick  bed  of  coals  its  passage  is  somewhat  retarded. 

"The  end  of  the  boiler  which  contains  the  man-hole  or  hand-hole 
should  be  set  one  inch  lower  than  the  other  end;  this  aids  the  flow 
of  the  water  and  sediment  toward  the  man-hole  through  which  it  can 
be  removed. 

"The  brick-work  is  closed  into  contact  with  the  shell  at  the  level 
of  the  center  of  the  upper  row  of  tubes;  this  prevents  the  gases  coming 
in  contact  with  the  plates  above  the  water-line. 

"A  safe  rule  is,  Never  expose  to  fire  or  gases  of  combustion  any 
part  of  the  shell  not  completely  covered  with  water. 

"The  brick  work  is  strengthened  by  buck-staves  held  together  by 
tie-rods.  The  buck-staves  are  of  wrought  iron,  channel  or  angle  irons. 

"In  the  matter  of  covering:  the  tops  of  boilers  and  other  portions 

57 


of  the  surface  not  in  contact  with  furnace  gases  should  be  covered  with 
some  non-conducting  substance  to  prevent  the  radiation  of  heat. 

"A  chimney  30  inches  in  diameter,  or  its  equivalent  area,  if  square 
(V06.9  sq.  in.),  with  a  height  of  90  feet,  will  be  ample  for  a  boiler  of 

"The  location  of  a  chimney  oftentimes  should  govern  the  height.  A 
chimney  should  have  height  enough  so  that  the  draught  should  not  be 
impaired  by  surrounding  objects." 


Q.  17.  (1900-01.)  Give  full  detailed  description  of  how  you  would 
conduct  a  boiler  test,  accompanied  by  a  report,  either  genuine  or  fic- 
titious, hand  firing. 

Give  rule  for  finding  the  factor  of  evaporation. 

Ans.  17.    Boiler  Tests — 

First— In  preparing  or  conducting  trials  of  steam  boilers  the  specific 
object  of  the  proposed  trial  should  be  clearly  defined  and  steadily  kept 
in  view. 

Second — Measure  and  record  the  dimensions,  positions,  etc.,  of  grate 
and  heating  surfaces,  flues,  chimneys,  proportion  of  air  space  in  the 
grate  surface,  kind  of  draught,  natural  or  forced. 

Third — Put  the  boiler  in  good  condition,  have  heating  surface  clean 
inside  and  out,  grate-bars  and  sides  of  furnace  free  from  clinkers, 
ashes  and  dust  removed  from  back  connections,  all  leaks  in  masonry 
stopped  and  all  obstructions  to  draught  removed.  That  the  damper 
will  open  to  full  extent,  and  that  it  will  close  when  it  is  desired. 

Fourth — Have  an  understanding  in  regard  to  the  character  of  coal 
to  be  used.  The  coal  must  be  dry;  if  wet  a  sample  must  be  dried 
carefully  and  the  per  cent  of  moisture  be  obtained,  to  correct  finally 
the  results  of  the  test. 

Fifth — In  all  important  tests  a  sample  of  coal  should  be  selected 
for  chemical  analysis. 

Sixth— Establish  the  correctness  of  all  apparatus  used  in  testing, 
weighing  or  measuring.  These  are:  (1)  scales  for  weighing;  (2) 
tanks,  or  water  meters  (water  meters  as  a  rule  should  be  used  only 
as  a  check  on  other  measurement) ;  for  accurate  work  the  water  should 
be  weighed  or  measured  in  a  tank;  (3)  thermometers  and  pyrometers 
for  taking  temperatures  of  air  and  steam,  feed  water,  waste  gases, 
etc.;  (4)  pressure  gauges,  draught  gauges,  etc. 

Seventh — Before  beginning  the  test,  the  boiler  and  chimney  should 
be  thoroughly  heated  to  their  usual  working  temperature.  If  the 
boiler  is  new,  it  should  be  in  continuous  use  at  least  one  week  before 
testing,  so  as  to  dry  the  mortar  thoroughly  and  heat  the  walls. 

Eighth— Before  beginning  a  test  all  superfluous  pipes  and  connec- 
tions should  be  disconnected  (including  the  blow-off),  or  stopped  with 
a  blank  flange,  unauthorized  opening  of  valves  should  be  guarded 
against.  If  an  injector  is  used,  it  must  receive  the  steam  directly 
from  the  boiler  under  test.  See  that  the  steam  pipe  is  so  arranged 
that  the  water  from  condensation  cannot  return  to  the  boiler.  If 
necessary,  it  must  be  trapped. 

Starting  and  Stopping  a  Test: 

A  test  should  last  at  least  ten  hours  of  continuous  running  and 
longer  if  practicable.  The  conditions  of  boiler  and  furnace  in  all  re- 
spects should  be,  as  nearly  as  possible,  the  same  as  at  the  beginning 
of  the  test.  The  steam  pressure  and  the  water-level  kept  the  same, 
the  fires  should  be  kept  as  near  as  possible  the  same;  in  fact,  all  con- 
ditions as  enumerated  kept  the  same. 

Standard  Method: 

Is  to  pull  the  fires,  the  steam  being  raised  to  the  proper  pressure, 
and  the  water-level  at  the  proper  point.  Close  damper  and  clean  ash- 
pit, etc.  Rebuild  with  weighed  wood  and  coal,  noting  the  time  and 

58 


all  conditions  of  water  and  steam.  At  the  end  of  the  test  remove  the 
fire  and  ashes  as  at  the  beginning  of  the  test,  and  make  notes  of  all 
conditions,  etc. 

An  Alternate  Method: 

Clean  the  fires,  and  note  all  conditions  as  in  the  standard  method, 
and  at  the  end  of  the  test  the  fires  should  be  burned  low,  as  at  the 
beginning  of  the  test  and  all  other  conditions  observed. 

During  the  test: 

"A."  Keep  all  conditions  uniform;  the  boiler  should  be  run  con- 
tinuously, without  stopping  for  meal-times  or  for  rise  or  fall  of  pres- 
sure of  steam  due  to  the  change  of  demand  for  steam.  The  draught 
being  adjusted  to  the  rate  of  evaporation  or  combustion  desired  before 
the  test  is  begun,  it  should  be  retained  constant  during  the  test  by 
means  of  the  damper. 

If  the  boiler  is  not  connected  to  the  same  steam  pipes  with  other 
boilers,  an  extra  outlet  for  steam  with  valve  in  same,  should  be  pro- 
vided, so  that  in  case  the  pressure  should  rise  to  that  at  which  the 
safety-valve  is  set,  it  may  be  reduced  to  the  desired  point  by  opening 
the  extra  outlet  without  checking  the  fires. 

If  the  boiler  is  connected  to  the  same  steam  pipe  with  other  boilers, 
the  safety-valve  on  the  boiler  being  tested  should  be  set  a  few  pounds 
higher  than  those  on  the  other  boilers,  so  that  in  case  of  a  rise  in 
pressure  the  other  boilers  will  blow  off,  and  the  pressure  be  reduced  by 
closing  their  dampers,  allowing  the  damper  on  the  boiler  to  remain 
open. 

Conditions  must  be  kept  uniform.  Should  (owing  to  the  character 
of  the  coal)  the  fires  need  cleaning  during  the  test,  the  time  and  all 
conditions  should  be  noted. 

Keeping  the  Records: 

The  coal  should  be  weighed  and  delivered  to  the  fireman  in  equal 
portions,  each  sufficient  for  one  hour's  run,  and  a  fresh  portion  should 
not  be  delivered  until  the  previous  one  has  all  been  fired,  noting  the 
time  required  to  consume  each  portion. 

It  is  desirable  that  at  the  same  time  that  the  amount  of  water  fed 
into  the  boiler  should  be  accurately  noted  and  recorded,  the  height  of 
the  water,  and  steam  pressure,  and  temperature  of  the  feed  water 
(average). 

By  thus  recording  the  amount  of  water  evaporated  by  each  suc- 
cessive portions  of  coal,  the  record  of  the  test  may  be  subdivided  into 
divisions,  if  desired,  and  at  the  end  of  the  test  to  discover  the  degree 
of  uniformity  of  combustion,  etc.,  at  the  different  stages  of  the  test. 

Priming  Tests. — Calorimeter  tests  should  be  made  to  ascertain  the 
percentage  of  moisture  in  the  steam  or  of  the  degree  of  superheating. 
At  least  ten  such  tests  should  be  made,  and  the  greatest  care  should 
be  taken  in  the  measurements  of  weights  and  temperature. 

Analysis  of  Gases.— Measurements  of  air  supply.  For  commercial 
purposes  are  not  necessary,  only  in  cases  of  scientific  research. 

These  are  the  measurements  of  the  air  supply,  the  determination 
of  its  contained  moisture,  the  measurement  and  analysis  of  the  flue 
gases,  the  determination  of  the  amount  of  heat  lost  by  radiation,  of 
the  amount  of  infiltration  of  air  through  the  setting,  the  direct  deter- 
mination by  calorimeter  experiments  of  the  absolute  heating  value 
of  the  fuel  and  (by  condensation  of  all  steam  made  in  the  boiler)  of 
the  total  heat  imparted  to  the  water. 

The  analysis  of  the  flue  gases  is  an  especially  valuable  method  of 
determining  the  relative  value  of  different  methods  of  firing  or  of 
different  kinds  of  furnaces. 

The  final  results  should  be  recorded  upon  a  properly  prepared 
blank  and  should  include  all  the  items  as  are  adapted  for  the  specific 
object  for  which  the  test  is  made. 

59 


REPORT  OP  ACTUAL  BOILER  TEST. 

Credited  to  "Mass.  No.  17,  Lowell,"  under  A.  S.  M.  E.  Code,  Stand- 
and  Wilcox  boiler,   with  B.   &  W.   regular   setting   plain 


used,  "Georges  Creek,"  Cumberland,  of  good  quality. 
Grate  surface,  67.6  sq.  ft. 
Water  heating  surface,  3,195  sq.  ft. 
Builders'  rating,  319.5  H.  P. 

TOTAL  QUANTITIES. 

1.  Date  of  trial,  May  30,  1899. 

2.  Duration  of  trial,  12  h. 

3.  Weight  of  coal  fired,  including  equivalent  wood,  14,966  pounds. 

4.  Percentage  of  moisture  in  coal,  3.1  per  cent. 

5.  Total  weight  dry  coal  consumed,  14,537  pounds. 

6.  Total  ash  and  refuse,  1,165  pounds. 

7.  Percentage  of  ash  and  refuse,  8.1  per  cent. 

8.  Total  weight  of  teed  water,  137,266  pounds. 

9.  Water  actually   evaporated,   corrected   for  moisture   in   steam, 
136,580  pounds. 

9a.     Factor  of  evaporation,  1.1913. 

10.  Equivalent  water  evaporated  into  dry  steam  from  and  at  212° 
F.,  162,708  pounds. 

HOURLY   QUANTITIES. 

11.  Dry  coal  consumed  per  hour,  1,211  pounds. 

12.  Dry  coal  per  sq.  ft.  grate  sur.,  per  hour,  17.9  pounds. 

13.  Water   evaporated   per   hour,   corrected   for  quality   of   steam, 
11,382  pounds. 

14.  Equivalent  evaporation  per  hour  from  and  at  212°  F.,  13,565 
pounds. 

15.  Equivalent  evaporation  per  hour  from  and  at  212°  F.,  per  sq. 
ft.  water  heat  surface,  4.25  pounds. 

AVERAGE  PRESSURE,  TEMPERATURE,  ETC. 

16.  Steam  pressure  by  gage,  91.9  pounds. 

17.  Temperature  of  feed  water  entering  boiler,  66°  F. 

18.  Temp,  of  escaping  gases  from  boiler,  386°  F. 

19.  Force  of  draught  between  boiler  and  damper,  .7  inch. 

20.  Percentage  of  moisture  in  steam,  5  per  cent. 

HORSE-POWER. 

21.  Horse-power  developed,  393  H.  P. 

22.  Builders'  rating  H.  P.,  319.5  H.  P. 

23.  Percentage  of  builders'  rating  developed,  excess,  23  per  cent. 

ECONOMIC  RESULTS. 

24.  Water  apparently  evaporated  under  actual  conditions  per  pound 
of  coal  as  fired,  9.15  pounds. 

25.  Equivalent  evaporation  from  and  at  212°  F.,  per  pound  of  coal 
as  fired,  10.85  pounds. 

26.  Equivalent  evaporation  from  and  at  212°  F.,  per  pound  of  dry 
coal,  11.19  pounds. 

27.  Equivalent   evaporation   from   and    at   212°    F.,    per   pound    of 
combustible,  12.17  pounds. 

EFFICIENCY. 

28  (assumed).    Calorific  value  of  dry  coal,  per  pound,  14,000  B.  T. 

29  (assumed).    Calorific  value  of  the  combustible  per  pound   15134 
B.  T.  U. 

60 


30.  Efficiency  of  boiler  (based  on  combustible),  77.6  per  cent. 

31.  Efficiency  of  the  boiler,  including  grate   (based  on  dry  coal), 
77.1  per  cent. 

COST   OF   EVAPORATION. 

32.  Cost  of  coal  per  ton  (2,240  pounds),  $3.40. 

33.  Cost  of  coal  required  to  evaporate  1,000  pounds  of  water  from 
and  at  212°  F.,  $0.136. 

The  Factor  of  Evaporation  is  the  difference  between  total  heat  of 
the  steam  at  the  observed  pressure  and  the  total  heat  of  the  feed  water 
at  the  observed  temperature,  divided  by  965.7. 
H-h 

Formula. =  Factor  of  evaporation. 

965.7 

H  =  the  total  heat  of  steam  at  the  observed  pressure, 
h  =  the  total  heat  of  feed  water  at  the  observed  temperature. 
965. 7  =  Latent  H.  U.  in  steam,  atmospheric  pressure. 


Q.  18.  (1900-01.)  If  your  test  should  show  that  you  were  evapo- 
rating fourteen  pounds  of  water  per  pound  of  coal,  or  six  pounds  of 
water  per  pound  of  coal,  would  you  regard  these  results  with  sus- 
picion, and  if  so,  why? 

Ans.  18.  In  the  common  forms  of  horizontal  tubular  land  boilers 
and  water-tube  boilers,  with  ample  horizontal  drums,  supplied  with 
water  free  from  substance  likely  to  cause  foaming,  the  moisture  in  the 
steam  does  not  usually  exceed  2  per  cent,  unless  the  boiler  is  over- 
driven or  the  water  level  carried  too  high. 

If,  in  an  evaporative  test,  it  was  found  that  fourteen  pounds  of 
water  was  being  evaporated  per  pound  of  coal,  the  result  would  be 
deemed  erroneous,  and  another  test  should  be  made,  subject  to  a  cal- 
orimetric  correction  for  the  unprecedented  amount  of  moisture  in  the 
steam.  There  are  not  sufficient  number  of  H.  U.  in  coal  to  evaporate 
14  pounds  of  water  per  pound  of  coal. 

On  the  other  hand,  there  might  be  losses  enough  to  reduce  the 
evaporation  down  to  6  Ibs.  per  pound  of  coal,  which  should  be  sought 
out  and  corrected  at  once. 


Q.  19.  (1900-01.)  What  is  the  difference  between  incrustation  and 
corrosion?  Define,  and  give  causes  for  each;  also  good  practice  with 
respect  to  prevention  of  each. 

Ans.  19.  Incrustation  is  due  to  the  presence  of  salts  in  the  feed 
water  (carbonates  and  sulphates  of  lime  and  magnesia  for  the  most 
part),  which  are  precipitated  when  the  water  is  heated,  and  form  hard 
deposits  upon  the  boiler  plates. 

Corrosion   (internal): 

In  marine  boilers  corrosion  is  generally  due  to  the  combined  action 
of  salt  water  and  air  when  under  steam,  and  when  not  under  steam 
to  the  combined  action  of  air  and  moisture  upon  •  the  unprotected 
surfaces  of  the  metal.  Other  deleterious  influences  are  at  work,  such 
as  the  corrosive  action  of  fatty  acids,  the  galvanic  action  of  copper  and 
brass,  and  the  inequalities  of  temperature;  these  latter  are  considered 
of  minor  importance. 

There  is  a  condition  where  a  boiler  is  badly  scaled,  the  high  tem- 
perature required  to  form  steam  predisposes  the  material  forming  the 
boiler  to  oxidize  rapidly. 

The  higher  the  temperature  at  which  iron  or  steel  is  kept,  the 
more  rapidly  it  oxidizes,  and  at  a  heat  of  600°  F.  it  soon  becomes  gran- 
ular and  brittle,  and  is  liable  to  bulge,  crack,  or  otherwise  give  way 
from  the  internal  pressure. 


As  a  preventative  from  scale  formation,  no  general  rule  can  be 
given;  it  is  only  through  a  careful  analysis  of  the  feed  water  that  the 
proper  antidote  is  found.  There  are  also  mechanical  appliances  known 
as  "live  steam  purifiers,"  by  the  use  of  which  the  impurities  are  pre- 
cipitated and  blown  off  before  reaching  the  boiler  proper.  When  the 
quantity  of  these  salts  is  not  very  large  (say  12  grains  per  gallon) 
scale  preventatives  may  be  found  effective.  The  chemical  preventa- 
tives  either  form  with  the  salts,  other  salts  soluble  in  hot  water,  or 
precipitate  them  in  the  form  of  soft  mud,  which  does  not  adhere  to 
the  plates,  and  can  be  blown  and  washed  out  at  times.  The  selection 
of  the  chemical  must  depend  upon  the  analysis  and  be  introduced 
regularly  with  the  feed. 

The  preventation  of  corrosion  in  a  boiler  is  a  subject  upon  which 
there  is  a  diversity  of  opinion.  A  boiler  in  use,  especially  where  re- 
turns from  heating  systems  and  otherwise,  or  where  pure  water  is 
used  which  has  a  tendency  toward  oxidization,  should  be  mixed  with 
a  certain  quantity  of  the  regular  service  water;  so_ne  propose  that  a 
boiler  after  cleaning  should  be  coated  with  some  compound  as  a  wash 
of  Portland  cement,  to  form  a  thin  hard  scale  to  protect  the  iron  from 
corrosive  action,  etc.,  etc. 

Boiler  laid  up  in  ordinary.  There  are  also  different  opinions  as  to 
which  is  the  proper  mode  of  procedure.  If  a  boiler  could  be  thoroughly 
cleansed  and  dried  and  kept  dry,  it  would  probably  be  as  good  a  way 
as  any,  but  in  connection  with  other  boilers,  there  is  more  or  less 
vapor  from  leaky  stop  valves  and  it  is  impossible  to  keep  all  parts  of 
the  boiler  in  a  dry  condition;  in  a  case  of  this  kind  I  would  suggest 
the  following  mode  of  procedure:  The  boiler  being  cleaned,  fill  with 
water  and  attach  a  ^-inch  steam  pipe  at  the  highest  point  and  keep 
the  boiler  under  pressure  at  all  times. 


Q.  20.  (1900-01.)  Would  you  regard  a  boiler  compound  offered  for 
universal  use  as  safe  to  use? 

What  may  be  considered  as  a  proper  preventative  under  any  and 
all  possible  conditions? 

Ans.  20.  A  boiler  compound  offered  for  universal  use  should  be 
condemned.  As  a  proper  preventative  to  be  used  at  all  times  and 
under  all  circumstances;  again  there  is  a  divergence  of  opinion. 
Generally  the  usual  substance  in  water  can  be  retained  in  soluble  form, 
or  be  precipitated  as  mud  by  using  caustic  soda  or  lime.  This  is 
especially  desirable  when  the  boilers  have  small  interior  spaces. 

Some  of  the  products  of  petroleum  are  highly  advocated  by  some, 
others  condemn  it.  Kerosene  (refined)  contains  more  or  less  of  the 
acids  used  in  the  refining  process,  yet,  still  it  is  recommended. 

Crude  petroleum  of  the  grade  known  as  rock  oil  is  probably  the 
best  product  of  the  petroleum  series;  the  grade  known  as  surface  oil 
should  be  barred,  as  it  contains  explosive  gases,  and  its  tarry  con- 
stituents are  apt  to  form  a  spongy  incrustation. 

Care  should  be  taken  on  opening  up  a  boiler  or  its  connections  at 
any  time  when  these  oils  have  been  used,  to  guard  against  explosions 
of  gas.  Lights,  cigars  and  pipes  should  be  left  in  the  background  and 
the  boiler  have  ample  time  to  air  out. 


Q.  24.  (1900-01.)  Assume  a  coal,  either  anthracite  or  bituminous, 
having  5  per  cent  moisture,  15  per  cent  refuse  and  the  balance  com- 
bustible capable  of  being  transformed  into  C  02  at  the  rate  of  14,544 
B.  T.  U.  per  pound.  Assume,  also,  that  one  horse-power  requires  2545 
heat  units  per  hour,  how  many  pounds  of  said  coal  will  be  required 

r  1200  horse-power  per  day  of  twelve  hours?  No  allowance  being 
made  for  banking  fires  and  waste  due  to  cleaning,  etc. 

62 


Ans.  24.     100— (15+5)  =80%  the  thermal  efficiency  of  the  coal. 

14544  X  80  =  11635.2  H.  U.  in  1  pound  of  the  coal. 

2545  X  1200  X  12  =  36,648,000  heat  units  required  under  conditions 
of  question  for  1200  H.  P.  for  a  run  of  12  hours. 

36,648,000-^11635.2  =  3149.7525  pounds  of  coal  required  per  day  of 
It  hours  for  1200  H.  P. 

This  amount  of  coal  consumed  of  course  looks  ridiculous,  and  the 
only  assumption  would  be  that  this  consumption  of  coal  was  based 
upon  a  theoretically  perfect  boiler,  and  an  engine  working  to  the 
absolute  zero  with  no  condensation  or  radiation  losses. 

Should  we  assume  that  this  engine  working  1200  H.  P.  for  day  of 
12  hours  had  a  thermal  efficiency  of  9  %; 

Then  will  36,648.000  H.  U.  equal  9  %  of  the  total  heat-units  in  the 
entire  amount  of  coal  consumed; 

Then  will  100  %,  or  total  consumption  of  coal  equal 
36,648,000  X  100 

=34,997  pounds  =17.49  tons  of  coal  of  2,000  Ibs.  per 

9 
ton,  required  for  1200  H.  P.  for  a  day  of  12  hours. 


Q.  27.  (1900-01.)  What  is  the  common  cause  of  smoke  and  what  is 
smoke  called  when  deposited  on  solid  bodies? 

Ans.  27.  The  ingredients  of  every  kind  of  fuel  commonly  used  may 
be  classed: 

First — Fixed  or  free  carbon,  which  is  left  in  the  form  of  charcoal 
or  coke  after  the  volatile  ingredients  of  the  fuel  have  been  distilled 
away. 

These  ingredients  burn  either  wholly  in  the  solid  states  (CtoCOa), 
or  part  in  the  solid  state  and  part  in  the  gaseous  state  (CO  +  O  = 
COs),  the  latter  part  being  first  dissolved  by  previously  formed  car- 
bonic acid  by  the  reaction,  C  O2  +  C  =  2  C  O. 

Carbonic  oxide,  C  O,  is  produced  when  the  supply  of  air  to  the  fire 
is  insufficient. 

Second— Hydro-carbons,  such  as  oleflant  gas,  pitch,  tar,  naphtha, 
etc.,  all  of  which  must  pass  into  the  gaseous  state  before  being  burned. 

If  mixed  on  their  first  issuing  from  amongst  the  burning  carbon 
with  a  large  quantity  of  hot  air,  these  inflammable  gases  are  com- 
pletely burned  with  a  transparent  blue  flame,  producing  carbonic  acid 
and  steam.  When  mixed  with  cold  air  they  are  apt  to  be  chilled  and 
pass  off  unburned.  When  raised  to  a  red  heat,  or  thereabouts,  before 
being  mixed  with  a  sufficient  quantity  of  air  for  perfect  combustion, 
they  disengage  carbon  in  fine  powder,  and  pass  to  the  condition  partly 
of  marsh  gas  and  partly  of  free  hydrogen,  and  the  higher  the  tem- 
perature the  greater  is  the  proportion  of  carbon  thus  disengaged. 

If  the  disengaged  carbon  is  cooled  below  the  temperature  of  igni- 
tion before  coming  in  contact  with  oxygen,  it  constitutes,  while  float- 
ing in  the  gas,  "smoke,"  and  when  deposited  on  solid  bodies  "soot." 


Q.  28.  (1900-01.)  What  occurs  when  the  disengaged  carbon  is 
maintained  at  the  temperature  of  ignition  and  supplied  with  sufficient 
oxygen  for  its  combustion? 

Ans.  28.  If  the  disengaged  carbon  is  maintained  at  the  tempera- 
ture of  ignition  and  supplied  with  oxygen  sufficient  for  its  combustion, 
it  burns  while  floating  in  the  inflammable  gas,  and  forms  red,  yellow 
or  white  flame.  The  flame  from  the  fuel  is  the  larger  the  more  slowly 
its  combustion  is  effected.  The  flame  itself  is  apt  to  be  chilled  by 
radiation  from  the  heating  surface  of  a  steam  boiler,  so  that  the  com- 
bustion is  not  completed,  and  part  of  the  gas  and  smoke  pass  off. 


Q  29  (1900-01.)  Does  the  flame  from  the  fuel  come  into  contact 
with' the  fire  sheets  of  a  horizontal  tubular  boiler,  or  the  outer  surface 
of  the  tubes  of  a  water  tube  boiler,  or  does  it  penetrate  into  the  tubes 
of  any  type  of  fire  tube  boilers? 

Ans  29.  In  the  use  of  wood  and  similar  class  of  fuels  there  is  no 
doubt  but  what  the  flame  comes  into  contact  with  the  fire  sheets  of 
the  boiler,  also  the  tubes  and  sometimes  the  stack. 

In  the  use  of  coal  (anthracite  and  semi-bituminous)  the  flame 
proper  is  hardly  long  enough  to  either  strike  the  sheets  or  penetrate 
the  tubes. 

In  mixing  the  hydro-carbons  with  the  necessary  amount  of  air  the 
product  of  perfect  combustion  takes  place  and  the  flame  encircles  the 
exposed  sheets,  penetrates  the  tubes,  especially  in  case  of  hard  driven 
toilers. 


Q.  30.  (1900-01.)  Can  secondary  combustion  take  place  in  the  back 
connections  or  flues? 

Ans.  30.  As  we  have  been  taught  that  combustion  is  the  chemical 
combination  of  the  constituents  of  the  fuel,  mostly  carbon  and  hydro- 
gen, with  the  oxygen  of  the  air.  The  nitrogen  in  the  air  remains  inert 
and  causes  loss  of  useful  effect  to  the  extent  of  the  heat  it  carries  off 
through  the  chimney. 

The  hydrogen  combines  with  the  proportional  part  of  oxygen  to 
form  water  which  passes  off  as  steam. 

The  carbon  combines  with  enough  oxygen  to  form  perfect  combus- 
tion, or  with  only  enough  to  form  imperfect  combustion. 

To  insure  perfect  combustion  the  sufficient  quantity  of  air  having 
been  admitted  and  properly  mixed  with  the  fuel  solid  and  gaseous,  and 
these,  the  air  and  combustible  gases  should  be  brought  together  and 
maintained  at  a  sufficiently  high  temperature.  The  hotter  the  elements 
the  greater  is  the  facility  of  good  combustion. 

If  perfect  combustion  ensues  there  is  no  need  of  secondary  combus- 
tion in  the  back  connections  and  flues. 

If  imperfect  combustion  takes  place,  then  a  portion  of  the  mixed 
air  and  gases  pass  off  unconsumed;  now,  unless  a  proportional  amount 
of  air  can  be  admitted  and  mixed  with  this  product,  and  the  tempera- 
ture raised  to  and  above  the  point  of  ignition  secondary  combustion 
will  not  take  place.  Secondary  combustion  is  desirable,  and  there 
has  been  various  devices  and  settings  introduced  at  various  times  to 
accomplish  this  result,  but  the  gain  has  not  warranted  their  contin- 
uance or  extended  use. 


Q.  31.  (1900-01.)  What  is  the  comparative  value  of  coal  and  oil  as 
fuel  from  the  evaporation  standpoint,  also  from  the  standpoint  of 
economy  in  cost? 

Ans.  31.  In  1892  there  were  reported  to  the  Engineers'  Club,  Phila- 
delphia, some  comparative  figures  from  tests  undertaken  to  ascertain 
the  relative  value  of  coal  and  oil  as  a  fuel. 

1  Ib.  anthracite  coal  evaporated  from  and  at  212°  F.,  9.70  Ibs.  water. 

1  Ib.  bituminous  coal  evaporated  from  and  at  212°  F.,  10.14  Ibs.  water. 

1  Ib.  oil  evaporated  from  and  212°  F.,  16.48  Ibs.  water. 

Taking  the  efficiency  of  the  bituminous  coal  as  a  basis  the  calorific 
energy  of  petroleum  is  more  than  60  %  greater  than  that  of  coal; 
whereas,  theoretically,  petroleum  exceeds  coal  only  45  %,  the  one  con- 
taining 14500  H.  U.  and  the  other  21000  H.  U. 

As  a  result  of  tests  made  by  the  Twin  City  Rapid  Transit  Company 
of  Minneapolis  and  St.  Paul,  showed  that  with  the  ordinary  Lima  oil 
weighing  6.6  Ibs.  per  gallon,  and  costing  2%  cents  per  gallon,  and  coal 
that  gave  an  evaporation  of  7%  Ibs.  of  water  per  pound  of  coal,  the 

64 


two  fuels  were  equally  economical  when  the  price  of  coal  was  $3.85 
per  ton  of  2,000  Ibs.  With  the  same  coal  at  $2.00  per  ton,  the  coal  was 
37  per  cent  more  economical  than  the  oil. 

With  the  coal  at  $4.85  per  ton  the  coal  was  20  %  more  expensive 
than  the  oil.  These  results  include  the  difference  in  the  cost  of  hand- 
ling the  coal,  ashes  and  oil. 


Q.  43.  (1900-01.)  Calculate  what  percentage  of  the  energy  contained 
in  the  fuel  is  utilized  in  a  steam  plant  consisting  of  a  boiler  which 
evaporates  eight  pounds  of  water  per  pound  of  fuel,  and  an  engine 
which  consumes  twenty-five  pounds  of  steam  per  IHP.  per  hour.  The 
fuel  contains  10,000  B.  T.  U.  per  pound. 

Ans.  43.  Pounds  of  coal  consumed  in  evaporating  1  Ib.  of  water  at 
200°  F.  into  steam  at  80  Ibs.  pressure  =  1  -^  8.  =  .125. 

Pounds  of  water  consumed  per  H.  P.,  25  Ibs. 

Pounds  of  coal  consumed  in  evaporating  25  Ibs.  of  water  at  200°  F. 
into  steam  at  80  Ibs.  pressure  =  25  X  .125  =  3.125  Ibs. 

Pounds  of  coal  consumed  per  hour  for  100  H.  P.  =  100  X  3.125  = 
312.5  Ibs. 

Heat  units  in  pound  of  fuel,  10,000  H.  U. 

Heat  units  contained  in  all  the  coal  consumed,  10,000  X  312.5  = 
3,125,000  H.  U. 

Mechanical  equivalent  for  one  H.  U.,  778  foot  pounds. 

Foot  pounds  of  work  stored  in  all  of  the  coal  consumed  each  hour 
=  778  X  3,125,000  =  2,431,250,000  ft.  Ibs. 

Foot  pounds  of  work  done  per  hour  by  each  I.  H.  P.,  33,000  X  60  = 
1,980,000  foot  pounds. 

Foot  pounds  of  work  per  hour  by  100  I.  H.  P.,  1,980,000  X  100  = 
198,000,000  foot  pounds. 

Losses  in  overcoming  friction  of  engine,  about  10  per  cent,  10  %  of 
198,000,000  =  19,800,000  foot  pounds. 

Total  foot  pounds  of  useful  work  at  the  engine  shaft  per  hour,  198,- 
000,000  —  19,800,000  =  172,000,000  foot  pounds. 

Foot  pounds  lost  per  hour,  2,431,250,000  —  172,000,000  =  2,259,250,000 
foot  pounds. 

Percentage  of  useful  work,  about  92-10  per  cent;  2,259,250000  — 
2,431,250,000  = 

Percentage  of  lost  work,  about  90  8-10  per  cent. 


Engines,  Indictaors,  Shafting,  Belting, 
Etc. 


Q.  5.  (1896-7.)  Are  liners  between  bearings  good  practice?  Why? 
How  properly  adjusted? 

Ans.  5.  Liners  between  bearing  boxes  are  good  practice.  1st,  they 
will  keep  dust  or  grit  from  getting  in  the  journals,  prevent  the  caps 
from  working  loose  and  prevent  rattling  of  the  boxes.  2d.  For  large 
engines  the  weight  of  the  top  box  and  cap  would  otherwise  increase  the 
friction.  Liners  are  adjusted  as  follows:  After  the  boxes  have  been 
carefully  fitted,  lead  wires  of  suitable  thickness  are  placed  between 
them.  The  cap  is  then  screwed  down  hard,  thereby  flattening  the  leads. 
They  are  taken  out  and  used  as  a  gage  to  fit  the  liners,  which  should  be 
a  trifle  thicker. 


Q.  18.  (1896-7.)  How  much  change  made  in  valve  travel  by  turning 
W  off  eccentric  [all  around]. 

Ans.  18.  None;  except  that  rod  length  will  have  to  be  altered  to 
maintain  position  of  valve. 


Q.  31.   (1896-7.)     What  is  the  proper  relation  between  cylinder  and 
steam  pipe  areas  for  automatic  cut-off  engines? 

Ans.  31.     Velocity  of  entering  steam  should  not  exceed  8,000  ft.  per 
minute. 

Area  of  cyl.  X  piston  speed 

Area  pipe  = — • 

Velocity  of  steam  in  pipe. 

A  constant,  6,000  ft.,  is  much  used  in  later  practice. 

Examples:     For  8,000  ft.,  area  pipe  =  area  piston  X  .075.     For  6,000 
ft.  area  pipe  —  area  piston  X  .10. 


eccentric?^896  ?-)     Wh&t  IS  meant  by  an&ular  advance  of  an  engine 

Ans.  34.    If  a  valve  has  neither  lap  nor  lead  and  delivers  steam  in 
te  usual  manner,  that  is,  over  or  around  the  ends  or  outside,  the 
eccentric  will  stand  at  an  angle  of  90°  with  the  crank  pin  either  in 
advance  of  the  crank  if  direct  connected,  or  behind  the  crank  if  indi- 


If  a  valve  has  lap  and  it  is  desired  that  the  valve 
open  the  port  for  steam  at  the  instant  the  crank  pin  moves  from 
the  dead  center,  or  if  the  valve  is  to  have  lead,  that  is,  to  open  for 

2SS  WH        S6  Cvank  is  on  tne  dead  center>  the  eccentric  must  be 
dyanced  on  the  shaft  in  the  direction  in  which  the  engine  is  to  run 
until  the  edge  of  the  valve  is  either  in  line  with  the  edge  of  the  steam 
„  ™   +11  thf  V.alVe  has  given  the  required  lead  opening,  the  engine 
-^ad   center.    Therefore,  the   angle   of  advance   or   the 
of  an  eccentric  is  the  angle   (beyond  ninety  degrees) 
scentric  is  advanced  or  pushed  around  on  the  shaft  to  bring 

66 


the  steam  edge  of  the  valve  in  line  with  the  steam  edge  of  the  port, 
if  the  valve  has  lap.  Or,  if  the  valve  has  lead  or  both  lap  and  lead,  it 
is  the  angle,  the  eccentric  is  advanced  (beyond  90°)  on  the  shaft  to 
give  an  opening  for  steam,  the  engine  in  either  case  being  on  the  dead 
center.  

Q.  37.  (1896-7.)  What  is  meant  by  angularity  of  the  connecting  rod 
of  an  engine? 

Ans.  37.  When  the  crank  pin  of  an  engine  stands  at  any  point  in 
its  path  of  motion,  except  when  on  the  dead  center,  the  connecting  rod 
forms  an  angle  with  the  line  of  centers,  this  is  termed  the  angularity 
of  the  connecting  rod.  The  angle  thus  formed  is  greatest  when  the 
piston  is  in  mid-position. 

By  the  line  of  centers  is  meant  a  straight  line  drawn  from  the 
center  of  the  crank  shaft  through  the  center  of  the  cylinder. 

The  shorter  the  connecting  rod  is  in  proportion  to  the  stroke  of  the 
piston,  the  greater  the  angularity  of  the  rod.  Or  the  angularity  of  the 
connecting  rod  may  be  described  as  the  equivalent,  but  not  actual, 
shortening  of  the  rod.  by  its  being  deviated  from  the  line  of  centers, 
which  it  is  at  all  times  except  when  the  crank  pin  is  on  the  dead 
centers.  

Q.  38.  (1896-7.)  Why  should  an  engine  be  given  compression? 
Name  all  the  reasons. 

Ans.  38.  Compression  is  given  to  an  engine — 1st.  To  furnish  a 
cushion  or  gradually  increasing  resistance  to  bring  the  reciprocating 
parts  of  an  engine  to  a  stop  at  the  end  of  the  stroke  and  change  the 
direction  of  the  thrust  upon  them  without  shock,  which  would  other- 
wise follow  upon  opening  the  steam  valve. 

2d.  If  obtained  by  early  closing  of  the  exhaust  valves  it  fills  the 
clearance  spaces  with  steam  that  would  otherwise  go  to  waste  by  being 
blown  into  the  condenser,  or  to  the  atmosphere. 

3d.  It  insures  quiet  and  smooth  running  of  the  engine,  taking  up 
the  lost  motion,  if  any,  without  jar. 

4th.  It  reheats  the  clearance  space  in  the  cylinder  [cylinder  head 
and  piston]  promoting  economy  by  thus  lessening  the  amount  of  in- 
ternal condensation,  a  very  important  item. 


Q.  39.  (1896-7.)  What  is  the  best  practice  and  most  accurate  way 
for  setting  Corliss  valves? 

Ans.  39.  Take  off  caps  of  valve  chambers;  marks  will  be  found; 
then  set  wrist  plate  so  that  central  line  on  wrist  plate  coincides  with 
line  on  stand ;  set  steam  valves  with  1-4"  lap  for  10"  diameter,  and  1-3" 
lap  for  32"  diameter;  for  intermediate  diameters  in  proportion.  Set 
exhaust  valve  1-16"  lap  for  10"  diameter,  and  1-8"  for  32"  diameter. 
Proportion  intermediate  diameters  on  non-condensing  engines;  on  con- 
densing engines  give  nearly  double  this. 

The  rods  connecting  steam  valve  arms  to  dash  pots  should  be  ad- 
justed by  turning  wrist  plate  to  extremes  of  travel  and  adjusting  the 
rod  so  that  when  it  is  down  as  far  as  it  will  go  the  steel  block  will 
just  clear  the  shoulder  of  the  hook.  Hook  the  engine  in  with  the  eccen- 
tric loose  on  shaft,  and  adjust  the  eccentric  rod  so  that  wrist  plate 
will  come  to  extremes  of  travel  as  indicated  by  lines.  Place  the  engine 
on  dead  center,  turn  eccentric  (in  direction  engine  runs)  until  steam 
valve  shows  from  1-32"  to  1-8"  lead,  owing  to  conditions;  turn  engine 
over  to  other  dead  center  and  note  lead  on  valve;  if  not  the  same, 
adjust  rod  from  wrist  plate  to  valve. 

To  adjust  rods  to  cut  off  from  governor:  Have  wrist  plate  at  one 
extreme  of  travel,  then  adjust  rod  connecting  the  opposite  cam  of 

67 


steam  valve  so  that  cam  will  clear  the  steel  tail  of  the  hook  about  1-32": 
turn  wrist  plate  to  other  extreme  and  adjust  the  rod  the  same  way.  To 
equalize  the  cut-off,  block  the  governor  up,  then  turn  engine  over  and 
note  where  cross-head  is  when  valve  is  tripped;  then  turn  on  over 
until  other  valve  is  tripped;  if  not  the  same,  adjust  one  or  the  other 
until  they  are  the  same.  Next,  start  the  engine  and  verify  results 
obtained  by  an  intelligent  application  of  the  indicator. 


Q.  44.  (1896-7.)  Is  there  a  gain  realized  in  practice  from  steam 
jacketed  cylinders?  If  so,  under  what  general  conditions? 

Ans.  44.  The  fact  that  a  gain  may  be  realized  by  steam  jacketing 
is  as  well  established  as  statements  to  the  contrary,  the  economy  in 
any  particular  case  depending  upon  design  and  conditions  attending 
their  application.  The  subject  is  quite  too  intricate  to  be  treated  satis- 
factorily in  an  abbreviated  manner.  Briefly  stated,  the  function  of  the 
steam  jacket  is  to  diminish  initial  condension  and  more  fully  main- 
tain the  temperature  of  the  expanding  steam  through  the  stroke.  For 
this  purpose  the  jacket  serves  as  a  medium  for  conveying  from  the 
boiler  a  reinforcement  of  heat  to  the  working  steam,  and  the  conditions 
to  be  observed,  both  in  design  and  management,  are  such  that  will  sub- 
serve this  end.  The  margin  of  gain  at  the  best  is  so  small  that  but 
a  slight  defect  in  design  or  operation  may  set  up  conditions  with  a 
tendency  just  the  reverse  of  those  which  the  use  of  a  jacket  implies. 
The  surplus  heat  stored  in  superheated  steam  performs  to  some  extent 
the  functions  usually  attributed  to  the  jacket,  consequently  the  benefit 
is  less  pronounced  when  such  is  used  within  the  cylinder  proper.  The 
gain  is  materially  greater  when  the  heads  as  well  as  the  body  of  the 
cylinder  are  steam  jacketed.  Efficiency  is  greater  for  long  stroke  and 
varies  somewhat  with  the  period  of  admission.  For  cylinders  in  series, 
compound,  etc.,  the  gain  is  greatest  in  the  high  pressure  cylinders,  since 
the  absorption  of  heat  during  the  exhaust  periods  is  not  lost  as  is  the 
case  of  the  simple  engine.  For  the  best  results  the  design  should 
insure  a  brisk  circulation  of  "live"  steam,  taken  directly  from  the 
boiler  through  all  parts  of  the  jacket,  and  ample  provision  must  be 
made  for  freeing  the  jacket  of  air  and  the  water  of  condensation,  the 
latter  to  drain  back  into  the  boiler. 


Q.  51.  (1896-7.)  What  should  be  the  ratio  between  steam  cylinder 
and  air  pump  capacity  for  a  condensing  engine? 

Ans.  51.  For  a  jet  condensing  engine,  the  single-acting  pump  should 
have  a  capacity  of  from  1-5  to  1-10  that  of  the  cylinder;  a  double-acting 
pump  would  have  1-8  to  1-16  the  cylinder  capacity.  For  surface  con- 
densing engine  the  single-acting  pump  would  have  a  capacity  of  1-10  to 
1-18  and  a  double-acting  from  1-15  to  1-25  that  of  the  cylinder.  In  both 
cases  the  proportions  depend  upon  the  terminal  pressure,  increasing 
with  same,  and  for  a  jet  condenser  varies  with  amount  of  injection 
used.  These  proportions  are  for  pumps  making  same  number  of  strokes 
3  the  engine.  If  the  speed  varies  from  that  of  main  engine  the  size  is 
so  calculated  as  to  give  same  relative  displacement  as  shown  above. 
When  an  engine  is  compounded  the  volume  of  the  low  pressure  cylinder 
alone  is  considered. 

Q.  52.     (1896-7.)     Is  brass  or  babbitt  best  for  bearings?    Elucidate. 

Ans.  52.  Except  in  case  of  bearings  where  the  pressure  is  excessive, 
where  pounding  and  jarring  is  liable  to  occur,  or  unusually  high  speeds 
are  attained,  babbitt  metal  must  be  given  the  preference  over  brass  as 
a  material  for  bearings.  Babbitt  bearings  can  be  the  more  easily  kept 
cool  and  there  is  less  liability  of  the  journal  cutting.  These  advantages 

68 


are  due  to  its  anti-friction  properties  and  to  the  fact  that  on  account 
of  its  softness  it  will  accommodate  itself  to  any  slight  irregularities  in 
the  surface  of  the  journal,  and,  should  the  shaft  get  slightly  out  of 
alignment  with  the  bearings,  the  metal  will  yield  where  the  pressure 
is  greatest  until  there  is  a  uniform  bearing  surface. 

Another  advantage  is  that  the  wear  of  the  shaft  will  be  less  rapid, 
and  when  the  bearings  become  worn  they  can  be  rebabbitted  at  a  slight 
expense.  To  give  best  results,  babbitt  metal  of  the  right  formula  or 
proportion  for  the  work  in  view  should  be  selected,  as  the  babbitts  are 
made  of  all  degrees  of  hardness  and  consistency.  It  is  our  opinion 
that  for  all  around  usefulness  there  is  no  bearing  metal  that  will 
approach  babbitt. 

Q.  55.  (1896-7.)  How  set  the  valves  on  a  medium  speed,  simple, 
non-condensing,  automatic,  Buckeye  engine? 

Ans.  55.  Under  the  conditions  of  the  question  it  is  understood  that 
the  regular  form  of  Buckeye  engine  is  meant,  and  it  will  be  necessary 
to  bear  in  mind  that  the  eccentric  follows  the  crank  instead  of  leading 
it,  as  in  the  ordinary  engine,  and  that  the  steam  chest  is  not  a  steam 
chest  as  commonly  understood,  but  is  an  exhaust  chamber  or  box;  that 
is,  the  cylinder  discharges  its  exhaust  steam  into  the  steam  chest  and 
takes  its  supply  of  steam  through  the  main  valve  direct  from  the  steam 
main.  There  are  two  valves  to  this  engine,  both  of  the  slide  valve 
type  and  both  flat,  and  their  functions  are  the  same  essentially  as  those 
of  the  common  slide  and  riding  cut-off.  The  riding  cut-off  in  this  case 
is  placed  on  the  inside  of  the  main  valve,  which  becomes  a  hollow  box 
or  chest  through  which  the  steam  passes. 

The  main  valve  has  ports  on  its  face  nearest  to  the  cylinder  which 
alternately  travel  over  the  usual  cylinder  ports  and  then  away  from 
same  far  enough  to  allow  the  end  of  the  valve  to  uncover  the  cylinder 
port  and  allow  the  steam  to  be  exhausted  from  the  cylinder  into  the 
steam  chest,  thus  enveloping  the  valve  and  keeping  it  at  all  times  in  a 
bath  of  hot  steam. 

There  is  a  separate  eccentric  for  each  valve,  one  for  the  main  valve 
fixed  to  the  shaft  and  the  other  for  the  riding  valve  or  cut-off,  operated 
by  a  loose  eccentric,  which  is  thrown  by  the  centrifugal  force  of  a 
weight  or  weights  around  the  shaft,  or  is,  in  other  words,  given  a 
variable  amount  of  angular  advance. 

The  main,  or  box  valve,  is  first  set  independently  of  the  cut-off  by 
putting  the  eccentric  90°  behind  the  crank,  plumbing  the  rocker  arm 
and  equalizing  the  compressions  just  as  would  be  done  with  the  leads 
on  a  common  slide  valve,  remembering  that  the  ends  of  the  valve  are 
exhaust  edges,  and  regulate  the  compression  or  exhaust  and  not  the 
steam  supply.  The  matter  of  equal  leads  in  this  engine  is  not  of  so 
much  importance. 

The  valves  and  cylinder  faces  are  marked  to  designate  the  position 
of  their  respective  ports,  which  are  in  themselves  invisible  from  the 
outside,  and  the  necessary  lead  is  given  by  advancing  the  eccentric 
toward  the  crank  or  in  the  direction  the  engine  will  run,  and  equalized 
in  the  usual  way. 

Thus  far  the  engine  is  the  same  as  a  common  slide  valve  type,  except 
that  the  steam  comes  in  through  the  valve  and  exhausts  out  of  and 
around  it.  The  ends  being  exhaust  edges,  the  eccentric  follows  instead 
of  leading  the  crank. 

Bearing  this  in  mind,  there  should  be  no  trouble  in  setting  the  valve. 
As  to  the  cut-off,  this  is  another  common  slide  valve,  much  smaller  and 
lighter  than  the  main  valve,  and  working  entirely  out  of  sight  and 
inside  of  the  main  valve,  being  rendered  visible  only,  and  then  with 
difficulty,  by  removing  the  back  plates  from  the  main  steam  chest  and 
withdrawing  the  balance  plates  found  therein.  This  valve  is  driven 
from  a  separate  eccentric,  fastened,  not  to  the  engine  shaft,  but  by 


links  to  the  governor  wheel  or  frame,  so  that  it  may  be  revolved  to  the 
extent  of  about  90°.  Its  normal  position  when  at  rest  is  coincident  with 
the  crank  of  the  engine,  and  is  held  there  by  the  tension  of  the  springs 
in  the  governor.  If  the  eccentric  is  set  in  the  same  direction  as  the 
main  crank,  that  is  all  that  is  necessary.  The  governor  is  then  set  out 
about  two-thirds  of  the  throw  of  its  arms  and  blocked,  and  the  valve  so 
adjusted  by  its  rods  and  stems  that,  as  shown  through  the  sight  holes, 
it  is  cutting  off  equally  on  each  end. 

It  has  been  assumed  that  the  experimenter  is  skilled  in  handling 
and  setting  the  ordinary  slide  valve.  Adjustments  in  the  case  of  each 
of  the  valves  of  this  engine  are  made  at  the  ends  of  the  eccentric  rods 
and  valve  stems,  as  usual,  keeping  in  mind  that  the  main  valve  acts 
just  exactly  opposite  to  the  manner  of  an  ordinary  slide. 

Compression  is  equalized  by  moving  the  main  valve  toward  the  end 
of  the  cylinder  showing  greatest  compression,  and  similarly  the  cut-off 
is  equalized  in  the  same  manner  by  moving  valve  toward  end  of  cylin- 
der showing  greatest  load.  After  the  cut-off  is  adjusted  and  equalized, 
any  adjustment  for  compression  equalization  on  the  main  valve  had 
better  be  made  on  main  eccentric  rod  and  not  on  valve  stem.  If  made 
on  the  latter  it  will  be  again  necessary  to  make  adjustments  for  cut- 
off. It  is  surmised  that  the  valves  and  parts  have  the  usual  shop  marks 
for  guidance  and  that  the  governor  wheel  and  its  parts  are  properly 
placed.  The  whole  of  the  work  should  be  finally  proved  and  corrected 
by  use  of  an  indicator. 


Q.  57.  (1896-7.)  If  we  get  an  initial  pressure  of  70  Ibs.  in  the  cylin- 
der and  the  compression  runs  up  to  35  Ibs.,  both  gage  pressure  will  the 
clearance  be  half  filled? 

Ans.  57.    Yes,  and  more  than  half  filled.    Thus: 
70  Ibs.  -f  15  =  85. 
35  Ibs. +  15  =  50. 

85H-2  =  42y2  Ibs.  clearance,  50%  filled,  or  42.5-^85X100  =  50%  or 
V2  filled;  hence,  under  the  above  conditions,  we  have  50  —  85  X  100  = 
58.8%,  which  is  more  than  half  filled. 


Q.  59      (1896-7  )     Will  the  slide  valve  cut  off  the  same  at  both  ends 
of  the  stroke  if  it  has  equal  lap  and  lead? 

hv  ^ft  59"  A*  ai\  engine  has  a  connecting  rod,  no.    If  crank  is  turned 
by  yoke  and   piston   rod,   yes. 


of  ?  pulley?1896'70     What  are  the  CaUS6S  °f  a  belt  runninS  to  one  si(le 

1;*  ^S?  carrying  Pullevs  n°t  being  parallel;  pulleys  not  in 
.        n  ;   °ne  6dge  °f  the  belt  beinS  stretched  more  than 


nhr.        n  nS  srece     more       an 

fdtrue  orb^n0ttCrTned  correctly:   PiHeys  not  being  bored  or 
ed  true,  or  belts  not  cut  square  before  lacing  or  glueing. 


Ans  62  The  valve  must  travel  in  each  direction  from  its  mid- 
travel  an  amount  equal  to  the  lap  and  width  of  the  port,  which  in  this 
case  is  the  sum  of  1  1-16"  and  15-16",  which  is  2",  thus  making  the 
valve  travel  4".  This  may  be  found  by  the  use  of  a  valve  model  or  any 
of  the  many  valve  diagrams  in  use. 


Q  68  (1896-7.)  If  an  engine  is  to  develop  300  H.P.  with  35  Ibs. 
m.e.p.  and  600  ft.  piston  speed  per  minute,  what  will  be  the  cylinder 
diameter?  Explain  how  found. 

Ans  68.  Rule  for  finding  the  required  diameter  of  cylinder  for  an 
engine  of  any  given  horse-power,  the  travel  of  piston  and  available 
pressure  being  given. 

Multiply  33,000  by  the  number  of  horse-power;  multiply  the  travel 
of  piston  in  feet  per  minute  by  the  available  pressure  in  the  cylinder; 
divide  the  first  product  by  the  second;  divide  this  quotient  by  the  deci- 
mal .7854.  The  square  root  of  this  last  quotient  will  be  the  required 
diameter  of  the  cylinder.  Example: 
33000  X  300-^-  (600  X  35) 

=  600. 

.7854 
V  600  =  24.4949"  diameter  required. 


Q.  72.  (1896-7.)  What  should  be  the  terminal  pressure  in  a  steam 
cylinder  of  40"  stroke,  2^%  clearance,  cut-off  at  1A  stroke,  initial  pres- 
sure 108  Ibs.  by  the  gage,  making  no  allowance  for  leakage,  condensa- 
tion or  wire  drawing,  and  assuming  15  Ibs.  atmospheric  pressure? 

Ans.  72.  The  terminal  pressure  will  be  in  the  same  proportion  to 
the  absolute  initial  pressure  as  the  volume  of  the  cylinder,  plus  the 
clearance  at  point  of  cut-off,  is  to  the  total  cylinder  volume  plus  the 
clearance. 

Stroke  40"  +  clearance   (2%%)   1"  =  41". 
Cut-off   ( 1A )   10"  +  clearance  1"  =  11". 
Total  pressure  (initial)   108  lbs.+ 15  lbs.=  123  Ibs. 
Thus:     P  :  123  :  :  11  in.  :  41  in. 

123  X  11 

=  33  Ibs.  absolute;   the  answer. 

41 


Q.  73.  (1896-7.)  What  is  the  distinction  between  indicated  and 
actual  steam  consumption  of  an  engine? 

Ans.  73.  An  indicator  diagram  does  not  show  the  actual  amount  of 
steam  used  by  an  engine  for  the  following  reasons: 

Leakage  of  steam  often  occurs  and  cylinder  condensation  is  inevi- 
table, yet  the  extent  of  these  losses  is  not  revealed  by  any  marked  effect 
upon  the  lines  of  a  diagram. 

The  measurement  of  steam  consumption  by  a  diagram  cannot,  there- 
fore, be  taken  to  show  actual  performance  without  allowance  for  these 
losses,  but  diagrams  taken  in  connection  with  feed-water  tests  will 
reveal  the  extent  of  the  losses  produced  by  leakage  and  cylinder  con- 
densation. These  losses  are  represented  by  that  part  of  the  feed  water 
consumption  which  remains  after  deducting  the  steam  computed  from 
the  diagram,  or  "accounted  for  by  the  indicator,"  as  it  is  termed.  The 
loss  by  condensation  is  nearly  constant  for  different  engines  working 
under  similar  conditions,  and  an  allowance  may  therefore  be  made  for 
its  amount.  The  other  leakage,  depending  upon  the  wearing  surfaces 
of  valves,  pistons  and  cylinder,  is  variable  in  different  cases. 


Q 


ZEUNER  VAI<VE  DIAGRAM 


Q.  76.  (1896-7.)  If  an  engine  is  to  run  160  revolutions  per  minute 
and  the  governor  pulley  256  revolutions  per  minute,  in  order  to  regulate 
speed  of  engine,  what  will  be  the  diameter  for  governor  shaft,  pulley,  if 
the  pulley  on  engine  shaft  is  8"  diameter? 

Ans.  76.     To  find  diameter  of  governor  shaft  pulley: 
Rule:      Multiply  the  number  of  revolutions  of  the  engine  by  the 
diameter  of  the  engine  shaft  pulley  and  divide  the  product  by  the 
number  of  revolutions  of  the  governor  pulley.    Example: 
160  (r.p.m.)  X  8  (inches) 

mi  5". 

256  r.p.m. 


Q.  80.  (1896-7.)  How  set  the  valves  on  a  Porter-Allen  simple 
engine? 

Ans.  80.  To  set  the  valves  of  the  Porter- Allen  engine:  There  are 
some  distinctive  features  peculiar  to  this  engine  «ot  found  in  other 
types  that  must  be  considered  when  setting  the  valves.  The  eccentric 
cannot  be  shifted  on  the  shaft,  as  in  most  other  types  of  engines;  it  is 
forged  and  turned  up  on  the  shaft;  it  is  in  the  same  position  on  the 
shaft  as  is  the  crank,  or,  in  other  words,  if  a  line  were  drawn  through 
the  center  of  the  cylinder,  crank-pin  and  shaft,  it  would  also  pass 
through  the  center  of  throw  of  the  eccentric.  The  object  of  this  is  to 
give  the  valves  the  same  relative  fast  and  slow  motion  as  that  of  the 
piston — when  the  piston  moves  slow,  the  valves  move  likewise,  and 
vice  versa. 

The  valves  are  four  in  number — two  admission  and  two  exhaust. 
They  are  placed  on  the  sides  of  the  cylinder,  steam  on  one  side,  exhaust 
on  the  other.  The  admission  valves  are  each  operated  by  a  separate 
valve  stem.  The  exhaust  valves  are  both  operated  by  a  single  stem 
upon  which  both  valves  are  mounted.  The  variable  cut-off  of  the  admis- 
sion valves  is  obtained  by  means  of  a  link  and  block  which  is  connected 
to  the  governor  in  such  a  manner  that  it  will  give  a  greater  or  less 
amount  of  opening  to  the  valves — the  amount  depending  on  the  load 
the  engine  is  to  carry,  varying  from  very  nearly  zero  to  56%  or  58% 
of  the  stroke.  The  exhaust  valves  are  operated  from  a  fixed  point  on 
the  link,  and  are  positive  in  their  motion,  regardless  of  the  point  of  cut- 
off of  the  admission  valves. 

From  the  foregoing  explanation  it  is  evident  that  the  adjustment  of 
the  valves  must  be  made  by  changing  their  relative  positions  on  the 
stems. 

To  set  the  valves,  the  first  thing  that  should  be  done  is  to  see  that 
the  valve  gear  is  in  perfect  harmony  with  the  governor  connections; 
this  is  done  by  turning  the  crank  to  one  or  the  other  centers,  but  not  on 
the  dead  center.  It  should  be  below  the  center  line.  A  shop  mark  will 
be  found  on  the  collar  of  the  shaft  and  another  on  the  corresponding  end 
of  the  upper  box  of  the  pillow  block.  When  these  two  marks  coincide 
the  crank  will  be  at  its  proper  place.  Then  raise  the  governor  to  its 
highest  position  and  let  it  down  again;  if  no  motion  is  given  to  the 
valves,  with  the  engine  in  this  position,  that  end  is  all  right.  If, 
however,  motion  should  be  given  to  the  valves  then  the  vertical  adjust- 
ment of  the  link  must  be  changed  until  there  is  no  motion  to  the  valves 
when  the  link  is  moved  and  the  crank  in  this  position. 

Vertical  adjustment  of  the  link  is  made  at  the  sustaining  or  sup- 
porting pin-boss  of  the  link  by  means  of  a  key  under  it  and  a  set  screw 
on  top  or  above  it.  The  construction  is  such  that  the  boss  pin  can  be 
raised  or  lowered  as  is  required.  It  may  so  happen  that  the  connecting 
rod  from  the  link-block  to  the  rocker-arm  may  have  to  be  lengthened  or 
shortened.  To  be  accurate  we  should  repeat  the  same  operation  on  the 
other  end.  This  operation  is  easily  understood  by  inspection  of  the 
engine. 

73 


faaving  found  the  operation  of  the  link  to  be  right,  turn  the  engine 
on  one  dead  center  and  raise  the  governor  to  the  highest  position  it  will 
go— this  will  bring  the  block  between  the  trunnions  of  the  link.  It 
must  be  securely  blocked  and  remain  in  this  position  while  the  admis- 
sion valves  are  being  set.  Proceed  to  set  the  valves  on  the  same  end  at 
which  the  piston  stands  (bearing  in  mind  that  the  valves  all  move 
toward  the  center  of  the  cylinder  to  open,  or  in  other  words,  the  valve 
on  the  head  end  moves  toward  the  crank  end  to  open,  and  vice  versa), 
giving  it  the  desirea  lead  which  will  depend  in  a  great  measure  on  the 
speed  and  size  of  engine,  the  kind  of  work  being  done,  the  steam  pres- 
sure and  sometimes  upon  the  fancy  of  the  operator.  The  construction 
of  the  valves  is  such  that  when  they  open  there  are  four  places  through 
which  steam  is  admitted.  The  amount  of  lead  is  generally  from  1/32" 
to  1/8",  depending  upon  the  above  named  conditions.  For  low  pressure 
cylinders  with  a  condenser  the  lead  is  as  much  as  3/16",  or  even  more. 
Next  turn  the  engine  on  the  other  center  and  give  the  valve  the 
same  lead.  This  being  done,  move  the  governor  to  its  lowest  position — 
where  it  is  when  the*engine  is  at  rest — it  will  then  be  seen  that  the  two 
admission  valves  have  moved  toward  the  crank  end  of  the  cylinder. 
This  motion  is  brought  about  by  moving  the  block  from  the  trunnions 
to  the  extremity  of  the  link  (or  from  one  extreme  to  the  other)  while 
the  crank  stands  on  dead  center.  The  motion  is  the  same  in  amount  on 
either  center  and  takes  place  in  the  same  direction — toward  the  crank 
end.  The  effect  is  to  cover  the  port  nearest  the  crack  and  enlarge  the 
opening  farthest  from  it.  The  lead  is  equal  to  the  earliest  point  of 
cut-off — it  is  gradually  diminished  on  the  crank  end  and  increased  at 
the  head  end.  This  is  in  the  same  proportion  as  the  steam  follows 
the  piston  in  either  stroke.  The  effect  is  to  equalize  the  points  of  cut- 
off at  either  end,  or  it  may  be  said  to  overcome  the  distortion  produced 
by  the  angularity  of  the  connecting  rod.  If  the  governor  is  blocked  up 
at  any  elevation  and  the  engine  turned  over  it  will  be  found  that  the 
openings  made  and  the  points  of  cut-off  will  be  the  same  for  each  stroke. 

The  exhaust  valves  have  a  positive  motion  and  are  both  mounted 
on  one  stem,  as  has  been  described.  To  set  the  exhaust  valves  make  a 
mark  on  the  guides  where  it  is  desired  that  compression  shall  take 
place,  or  the  point  where  the  valve  shall  close  when  the  piston  has 
nearly  completed  its  stroke.  Turn  the  engine  until  the  crosshead 
comes  to  this  mark  and  set  the  edge  of  the  valve  on  line  with  edge  of 
the  port  so  that  a  slight  movement  of  the  crosshead  either  way  will 
open  or  close  the  valve.  Turn  the  engine  to  the  other  end  and  repeat 
the  operation. 

The  point  at  which  the  exhaust  port  should  close  is  from  2.5"  to  5" 
from  end  of  stroke.  With  condensing  engines  even  more  than  this  is 
sometimes  given,  the  amount  varying  somewhat  with  the  speed,  etc., 
but  enough  must  be  given  to  make  the  engine  run  quietly. 

The  exhaust  valves  can  be  set  while  turning  the  engine  to  set  the 
admission  valves;  this  will  avoid  turning  the  engine  so  often.  The 
valves  are  all  held  in  position  on  the  stems  by  two  pairs  of  nuts  which 
must  be  securely  locked  and  leave  sufficient  play  so  that  the  valve  can 
adjust  itself  sideways  but  at  the  same  time  have  no  end  play. 

The  valves  a.*e  held  in  position  by  what  are  known  as  pressure 
plates.  These  plates  must  be  sufficiently  close  to  make  them  steam 
tight,  yet  not  so  tight  as  to  cause  unnecessary  friction. 

If  the  valve  setting  has  been  done  by  one  not  thoroughly  familiar 
with  this  type  of  engine  it  should  be  turned  one  revolution  and  the 
valves  and  gear  carefully  inspected  to  see  that  all  parts  move  with 
freedom  and  accuracy;  then  put  on  the  steam  chest  covers,  warm  the 
cylinder,  start  the  engine  and  apply  the  indicator. 

If  the  indicator  diagrams  should  show  that  more  or  less  lead  is 
needed  on  the  admission  valves  the  adjustment  can  be  made  without 
removing  the  steam  chest  covers  by  lengthening  or  shortening  the 

74 


valve  stems  as  the  case  may  be; — but  this  cannot  be  done  on  the 
exhaust  side,  due  to  the  valves  being  mounted  on  one  stem,  unless 
both  valves  should  require  to  be  moved  in  the  same  direction,  then  the 
stem  could  be  made  longer  or  shorter  without  removing  the  covers. 


Q.  86.  (1896-7.)  For  ordinary  continuous  duty,  how  many  r.p.m. 
would  you  run  a  150  HP  simple,  Corliss  engine?  Why? 

Ans.  86.  We  consider  the  proper  size  for  a  Corliss  engine  to  develop 
150  HP  to  be  as  follows:  16"  X.42"  at  78  r.p.m.  100  Ibs.  B.P.,  cut-off 
1/5  stroke. 

By  consulting  various  builders'  catalogues,  the  rated  r.p.m.  for  this 
size  engine  is  75  to  80.  This  we  consider  a  fair  rate.  Excellent  results, 
as  far  as  engine  performance  and  steam  economy,  caii  be -attained  at  a 
higher  rotative  speed,  as  with  85  or  90  r.p.m.,  above  this  the  steam  lines 
are  not  sharp,  and  the  wear  upon  valve  gear  becomes  very  marked; 
also  the  silent  operation  of  the  engine  apparatus  is  affected.  Corliss 
engines  are  running  nicely  at  speeds  as  high  as  150  or  180  r.p.m.  The 
reason  for  keeping  the  speed  down  is  that  the  parts,  especially  the  valve 
mechanism  and  valves,  are  more  easily  kept  in  repair,  a  minimum  of 
noise  is  produced  by  the  engine  and  dash  pots  are  given  time  to  do 
effective  and  perfect  work. 


Q.  87.  (1896-7.)  A  common  D  slide  valve  travels  enough  to  give 
full  port  opening,  outside  lap  is  1  1/16",  inside  lap  is  3/16"  and  width  of 
port  is  I",  when  1-16"  lead  is  given,  show  by  a  valve  diagram  the  points 
of  cut-off,  release  and  compression,  ignoring  distortion  caused  by  con- 
necting or  eccentric  rod. 

Ans.  87.  Use  the  Zeuner  valve  diagram  (see  diagram)  and  proceed 
as  follows:  Describe  a  circle,  with  a  radius  OA  equal  to  the  half 
travel  of  the  valve.  From  O  measure  off  OB  equal  to  the  outside  lap, 
and  BC  equal  to  the  lead.  When  the  crank  pin  occupies  the  dead  center 
A,  the  valve  has  already  moved  to  the  right  of  its  central  position  by 
the  space  OB  plus  BC.  From  C  erect  the  perpendicular  CB  and  join  OB. 
Then  will  OB  be  the  position  occupied  by  the  line  joining  the  center  of 
the  eccentric  with  the  center  of  the  crank-shaft  at  the  commencement 
of  the  stroke.  On  the  line  OB  as  diameter  describe  the  circle  OCE, 
then  any  chords,  as  Oe,  OB,  Oe,  will  represent  the  spaces  traveled  by 
the  valve  from  its  central  position  when  the  crank-pin  occupies  respec- 
tively the  positions  opposite  to  D,  E  and  F.  Before  the  port  is  opened 
at  all  the  valve  must  have  moved  from  its  central  position  by  an  amount 
equal  to  the  lap  OB.  Hence,  to  obtain  the  space  by  which  the  port  is 
opened,  subtract  from  each  of  the  arcs  Oe,  OE,  etc.,  a  length  equal  to 
OB.  This  is  represented  graphically  by  describing  from  center  O  a 
circle  with  radius  equal  to  the  lap  OB,  then  the  spaces  fe,  gE,  etc., 
intercepted  between  the  circumference  of  the  lap-circle  Bfe  and  the 
valve  circle  OCE,  will  give  the  extent  to  which  the  steam-port  is  opened. 
At  the  point  k,  at  which  the  chord  Ok  is  common  to  both  valve  and  lap 
circles,  it  is  evident  that  the  valve  has  moved  to  the  right  by  the 
amount  of  the  lap,  and  is  consequently  just  on  the  point  of  opening  the 
steam-port.  Hence  the  steam  is  admitted  before  commencement  of  the 
stroke,  when  the  crank  occupies  the  position  OH,  and  while  the  portion 
HA  of  the  revolution  still  remains  to  be  accomplished.  When  the 
crank-pin  reaches  the  position  A,  that  is  to  say,  at  the  commencement 
of  the  stroke,  the  port  is  already  opened  by  the  space  OC  less  OB  equals 
BC,  called  the  lead.  From  this  point  forward  until  the  crank  occupies 
the  position  OE,  the  port  continues  open,  but  when  the  crank  is  at  OE 
the  valve  has  reached  the  furthest  limit  of  its  travel  to  the  right,  and 
then  commences  to  return,  until  when  in  the  position  OF  the  edge  of  the 

75 


K 


.  2. 


Fig.  1. 


I        I 


TU- 


VALVE  DIAGRAM. 
76 


valve  just  covers  the  steam-port,  as  is  shown  by  the  chord  Oe  being 
again  common  to  both  lap  and  valve  circles.  Hence  when  the  crank 
occupies  the  position  OF,  the  cut-off  takes  place  and  the  steam  com- 
mences to  expand,  and  continues  to  do  so  until  the  exhaust  opens. 
When  the  line  joining  the  centers  of  the  eccentric  and  crank  shaft 
occupies  the  position  opposite  to  OG  at  right  angles  to  the  line  of  dead 
centers,  the  crank  is  in  the  line  OP  at  right  angles  to  OB,  and  as  OP 
does  not  intersect  either  valve-circle,  the  valve  occupies  its  central  posi- 
tion, and  consequently  closes  the  port  by  the  amount  of  the  inside  lap. 
The  crank  must  therefore  move  through  such  an  angular  distance  that 
its  line  of  direction  OQ  must  intercept  a  chord  on  the  valve  circle  OK 
equal  in  length  to  the  inside  lap  before  the  port  can  be  opened  to  the 
exhaust.  This  point  is  ascertained  by  drawing  a  circle  from  center  O 
with  a  radius  equal  to  the  inside  lap;  this  is  the  small  inner  circle 
in  the  figure.  Where  this  circle  intersects  the  valve  circles,  we  get  the 
points  which  show  the  positions  of  the  crank  when  the  exhaust  opens 
and  closes.  Thus  at  Q  the  valve  opens'  the  exhaust  on  the  side  of  the 
piston  which  we  have  been  considering,  while  at  R  the  exhaust  closes 
and  compression  commences  and  continues  till  the  fresh  steam  is  ad- 
mitted at  H.  To  resume  the  operations:  Steam  admitted  before  the 
commencement  of  the  stroke  at  H.  At  the  dead  center  A  the  valve  is 
already  opened  by  the  amount  BC  equal  to  1/16".  At  E  the  port  is 
fully  opened,  and  the  valve  has  reached  one  end  of  its  travel,  or  2  1/16" 
from  its  central  position.  At  F  steam  is  cut  off,  consequently  admission 
lasted  from  H  to  F.  At  P  valve  occupies  central  position,  and  ports 
are  closed  to  both  steam  and  exhaust.  At  Q  exhaust  opened  and  expan- 
sion lasted  from  F  to  Q.  At  K  exhaust  opened  to  maximum  extent,  and 
valve  reached  the  other  end  of  its  travel.  At  R  exhaust  closed,  and  com- 
pression begins  and  continues  till  the  fresh  steam  is  admitted  at  H. 


EXPLANATION  OF  BILGRAM  DIAGRAM. 

Fig.  1  is  the  diagram  proper;  the  circle  ABC  represents  the  path  of 
the  center  of  the  eccentric.  The  circle  of  1"  radius  drawn  about  the 
center  O  is  the  port  circle.  The  large  and  small  circles  drawn  about 
center  Q  represent  the  steam  and  exhaust  laps  respectively;  d.h.f. 
represent  the  points  of  cut-off,  compression  and  release  for  the  full  line 
card  projected  to  d'h'f  in  Fig.  2.  Also  i,e,g  represent  corresponding 
points  projected  to  i',e',g',  for  the  dotted  line  diagram,  also  shown  in 
Fig.  2. 

The  dimensioned  space  marked  1/16"  equals  lead  of  valves.  The 
circle  ABC  is  also  used  to  represent  the  crank  circle  to  any  scale,  as 
the  length  of  the  crank  is  not  given.  No  attempt  has  been  made  to 
draw  theoretical  curves  in  Fig.  2,  the  object  of  this  diagram  being  a 
graphic  explanation  of  Fig.  1. 


THE  SWEET  VALVE  DIAGRAM— EXPLANATION. 

Lay  off  the  horizontal  line  AC;  draw  the  circle  ABCD  equal  in  the 
diameter  to  the  width  of  the  port  and  outside  lap  of  valve;  this  gives  the 
valve  travel  in  full  measurement  (being  in  this  case  4%");  draw  the 
circle  EF.  This  is  equal  in  diameter  to  twice  the  width  of  the  outside 
lap  of  the  valve  (in  this  case  2%").  Around  the  same  center  draw  the 
circle  G,  which  is  equal  in  diameter  to  twice  the  width  of  the  inside  lap 
(in  this- case %").  Draw  the  two  small  circles  at  A  and  C  equal  in  diame- 
ter to  twice  the  lead  (%"  in  this  case).  Lay  off  the  line  Ca  to  touch  the 
lead  circle  and  outside  lap  circle.  Where  it  intersects  the  valve  circle 
at  a,  is  the  point  of  cut-off.  Lay  off  the  two  lines  be  ana  cb  parallel  to  Ca, 
touching  on  the  inside  lap  circle  where  these  two  lines  intersect  the 

77 


SWEET  VAI,VE  DIAGRAM. 
78 


valve  circle  at  b  and  c  is  the  point  of  release  and  compression,  b  being 
release  and  c  compression. 

Above  the  diagram  are  indicator  cards.  Below  is  a  scale  of  24" 
stroke,  showing  theoretically  all  the  points  at  which  the  events  would 
take  place,  by  following  the  heavy  and  dotted  lines  and  their  letters. 
The  theoretical  result  of  the  indicator  card  here  given  is  seldom  if  ever 
obtained  in  actual  practice. 


Q.  88.  How  many  ropes  required  to  transmit  85  HP  at  a  rope  speed 
of  2,500  ft.  per  minute,  breaking  strain  of  ropes  equals  9,000  Ibs.,  work- 
ing strain  1%  per  cent  of  the  breaking  strain? 

Ans.  88.  Under  the  conditions  imposed  the  number  of  ropes  required 
will  be  ten  (10)  and  may  be  determined  as  follows: 

Reduce  the  horse-power  to  foot  pounds,  then  33,000  X  85  =  2,805,000 
ft.  Ibs.,  which  divided  by  2,500  ft.  (rope  speed)  gives  1,122  Ibs:.  as  the 
gross  pull  on  all  the  ropes;  1%%  of  9,000  Ibs.,  or  112.5  Ibs.,  is  the  allow- 
able strain  for  any  one  rope,  and  divided  into  1,122,  the  total  strain, 
will  give  9.97  +,  or  say  in  even  numbers  10  ropes. 


Q.  92.     (1896-7.)     Name  three  remedies  for  cylinder  condensation. 
Ans.  92.     1.     Compounding. 

2.  Steam  Jacketing. 

3.  superheating. 


Q.  95.  (1896-7.)  Are  two-cylinder,  non-condensing,  compound,  high 
speed  engines  of  less  than  100  HP  economical? 

Ans.  95.  Tests  made  by  practical  engineers  have  demonstrated  that 
the  steam  consumption  of  the  type  of  engine  named  in  the  question  is 
not  less  than  26  Ibs.  per  HP  per  hour,  and  very  seldom  as  low  as  the 
figure  given.  The  first  cost  of  such  an  engine  is  out  of  proportion  to 
the  slight  gain  in  economy.  The  cost  of  maintenance,  the  increased 
friction  and  the  space  required  make  such  an  engine  undesirable. — 
Note:  Under  proper  conditions  of  pressure,  speed  or  load  this  type  of 
engine  may  be  economical  of  steam,  but  it  is  the  solitary  exception  to  a 
broad,  general  rule  to  the  contrary. 


Q.  96.  (1896-7.)  What  is  the  maximum  allowable  pressure  per 
square  inch  on  a  crank  pin?  . 

Ans.  96.  This  differs  with  different  authorities,  some  claiming  that 
the  maximum  pressure  should  not  exceed  400  Ibs.  to  the  square  inch  of 
projected  area,  equal  to  diameter  X  length  of  pin,  while  others  claim 
as  high  as  800  to  1,200  Ibs.,  to  the  square  inch.  Would  consider  that  for 
iron  pins  a  pressure  of  500  Ibs.  and  for  steel  800  Ibs.  to  the  inch  not 
excessive,  especially  as  the  load  is  intermittent.  It  is  doubtful  whether 
the  pins  would  stand  a  constant  pressure  like  that  given.  Less  pres- 
sure should  be  used  with  high  speeds. 

Under  above  conditions  lubrication  becomes  an  easy  problem. 


Q.  98.  (1896-7.)  How  large  is  the  low  pressure  cylinder  of  a  com- 
pound engine  as  compared  with  the  cylinder  of  a  simple  engine  doing 
equal  work  with  the  same  speed  and  pressure? 

79 


Ans.  98.  If  the  initial  and  terminal  pressures  are  to  be  the  same,  the 
ratio  of  expansion  must  be  the  same  in  both  cases,  and  if  the  work  done 
is  equal,  then  the  cylinders  must  be  of  the  same  size.  To  illustrate: 
Take  case  of  a  simple  engine  20"  X  36",  100  r.p.m.,  135  Ibs.  steam, 
exhausting  at  15  pounds — both  absolute.  Then,  135  -=-  15  =  9,  or  ratio 
of  expansion  with  back  and  terminal  pressures  equal.  The  m.e.p.  will 
be  32.95  Ibs.  and  the  HP  314  X  32.95  X  600  -=-  33,000  =  187.8. 

If  a  compound  is  to  be  used,  the  low  pressure  cyl.  will  be  20  X  36, 
the  ratio  for  each  cylinder  will  be  3  and  the  terminal  in  the  high  pres- 
sure cylinder  will  be  135  -f-  3  =  45  Ibs.  With  equal  back  pressure  the 
m.e.p.  will  be  49.45  Ibs.  and  the  HP  104  X  49.45  X  600  -=-  33,000  =  93.5. 
Assuming  the  initial  pressure  in  low  pressure  cylinder  to  be  45,  the 
terminal  will  be  45  ~  3  =  15  and  the  m.e.p.  16.48;  then  314  +  16.48  X 
600^33,000  =  94  HP. 

93.5  +  94  =  187.5  power  for  compound  engine,  same  as  for  simple. 


Q.  99.     (1896-7.)     What  is  the  safe  rim  speed  for  an  18  ft.  cast,  split, 
band-wheel,  5"  thickness  of  rim,  and  36"  face? 
Ans.  99.    Holmes,  pp.  247  and  161,  gives: 

W  X  TS  Xr  X  .00034 
R.P.M.  =  - 

2  Xir 

W  =  weight  of  rim  in  Ibs. 
TS  =  safe  strength  of  iron  =  700  Ibs. 

r  =  radius  =  9  ft. 
.00034  =  a  constant. 

ir=  3.1416;  then  substituting 
W  =  18  X  3.1416  X  12  X  5  X  36  X  .2604  =  31807   Ibs.    and 

31807  X  700  X  9  X  .00034 
R.P.M.  =  _ 

6.2832 
and  R.P.M.  =  104  =  97  ft.  per  second. 

C  the  mean  radius,  something  less  than  9  ft.,  and  the  mean  weight 
had  been  used  the  result  would  have  2%  to  5%  less  than  above 

Theoretically   leaving  out  of  consideration  the  rigidity  of  construc- 
of  Us  weight  a  Wheel>  aS  resards  centrifugal  force  is  independent 


gH  8?2™™y«S  at  5°°°  ft  per  min"  an  18  ft-  wheel  may  run 
.  and  at  6000  ft.  per  minute  may  run  105  R.P.M.,  but  does  not 
recommend  such  high  rim  speeds  for  cast  iron  wheels. 


norse 


abolf  10^UHRatiC    CUt"°ff   Single    cylinder    non-condensing    engines    of 
practice?0^16  expansion  condensing  engines  of  about  500  HP.,  in  good 
Ans.  23.     (a)  About  25  to  30.     (b)  About  12  to  15. 


conden^ng  engine   has   a   ter- 


1u 

recommend  to'better  adapt  ^  T  Us'  wor'k?     "  °f  ^  ^^  W°Uld  Y°U 
Ans.  24.     Compound  the  engine  by  adding  a  low  pressure  cylinder. 


80 


Q.  28.  (1897-8.)  An  engine  has  a  twelve  (12")  inch  stroke.  Its  con- 
necting rod  is  three  (3)  feet  long.  It  is  running  at  six  hundred  (600) 
revolutions  per  minute.  The  weight  of  its  piston,  piston-rod  and  cross- 
head  is  250  Ibs.  Draw  to  scale  a  diagram  such  that  horizontal  distances 
shall  represent  piston-positions  and  vertical  distances,  piston-velocities 
at  the  respective  positions,  neglecting  the  angularity  of  the  connecting- 
rod. 

Ans.  28.  Lay  off  horizontally,  to  a  convenient  scale,  the  larger  the 
better,  a  line  A,  C,  Fig.  2,  to  represent  the  stroke  of  the  piston.  Draw 
upon  this  line  a  semi-circle  A,  B,  C.  Then  will  points  upon  the  horizon- 
tal line  represent  piston  positions  and  the  vertical  distances  above  said 
points  to  the  semi-circle  will  represent  piston  velocities  at  the  corre- 
sponding positions  of  the  piston. 

The  radius  B,  D,  of  the  semi-circle  represents  the  velocity  of  the 
crank-pin,  and  the  vertical  lines  at  other  points  of  the  stroke  velocities 
proportional  to  their  lengths. 


FlG.  2 


Q.  29.  (1897-8.)  In  the  engine  of  Q.  28  what  is  the  approximate 
velocities  of  the  piston  at  positions  one  and  one-half  (1-5")  and  two 
(2")  inches  from  the  commencement  of  its  stroke?  Explain  how  you 
find  this  by  the  diagram  of  Ans.  28. 

Ans.  29.  In  Fig.  2,  the  scale  is  V±"  to  the  inch.  The  radius  of  the 
semi-circle  is  therefore  1.5".  The  vertical  line  b,  b,  at  the  point  which 
represents  a  position  1.5"  from  the  commencement  of  the  stroke,  is 
.992"  long.  The  velocity,  therefore,  at  this  point  is  found  by  the  pro- 
portion 1.5  :  .992  :  :  31.4  :  answer.  By  multiplying  31.4  by  .992  and 
dividing  by  1.5  we  find  the  velocity  at  this  point  to  be  20.77  ft.  per 
second. 

The  vertical  line,  c,  c,  at  the  position  corresponding  to  2"  from  the 
commencement  of  the  stroke  is  1.118"  long.  Calculating  as  above  the 
proportion  would  be  1.5  :  1.118  :  :  31.5  :  23.4  Therefore  the  velocity  at 
2"  from  the  commencement  of  the  stroke  is  23.4  ft.  per  second. 


Q.  30.  (1897-8.)  With  the  engine  of  Q.  28  what  is  the  approximate 
average  force  due  to  the  inertia  of  the  piston,  piston-rod,  and  cross-head, 
while  the  piston  is  passing  from  a  position' one  and  one-half  (1.5")  to 
a  position  two  (2")  inches  from  the  commencement  of  its  stroke?  Solve 
by  the  use  of  the  rule  of  Ans.  27  and  the  diagram  of  Ans.  28,  and  explain 
fully  the  method  of  calculation. 

Ans.  30.  The  weight  of  the  parts  is  250  Ibs.  The  velocity  at  1.5"  is 
20.77  ft.  per  second.  The  velocity  at  2"  is  23.4  ft.  per  second.  There- 
Si 


fore,  according  to  the  rule  of  Ans.  27,  the  work  done  upon  the  parts  to 
change  their  velocity  from  one  value  to  the  other  is  250  X  [(23.4)2 — 
(20.77)2]  ~  by  64.4,  that  is  250  X  (547.56  —  431.4)  -=-  64.4  =  250  X  116.16 
-=-  64.4  =  541  foot-pounds  of  work  done  upon  the  reciprocating  parts  to 
increase  their  velocity  in  W  travel.  Now  this  work  divided  by  the 
distance  will  give  the  average  force  exerted.  Thus  451  foot-pounds  of 
work  divided  by  one  twenty-fourth  of  a  foot  equals  10,824  Ibs.  as  the 
average  force  due  to  inertia. 

The  change  of  angularity  of  the  connecting  rod  may  be  taken  into 
account  by  multiplying  the  result  by  one  plus  the  ratio  between  the 
crank  and  connecting  rod,  that  is,  in  this  instance,  by  one  and  one-sixth, 
[f  the  calculation  had  been  made  on  the  latter  half  of  the  stroke,  the 
correction  would  have  been  made  by  multiplying  by  unity,  minus  the 
ratio  of  the  crank  to  the  connecting  rod;  in  this  instance  by  7/6. 


Q.  34.  (1897-8.)  In  the  engine  of  Q.  28,  if  the  steam  is  suddenly 
shut  off  while  the  engine  is  running  at  full  speed  and  running  over, 
what  force  would  be  necessary  to  keep  the  cross-head  from  pressing 
against  the  upper  guide  when  the  piston  is  one  inch  from  the  com- 
mencement of  its  forward  stroke?  Explain  method  of  calculation. 

Ans.  34.  As  the  steam  is  shut  off  we  only  have  to  deal  with  the 
forces  of  inertia.  By  referring  to  the  diagram  Fig.  3,  we  find  that  the 
force  at  this  point,  1"  from  the  commencement  of  the  stroke,  is  14,000 
Ibs.  Now  draw  a  line  a,  b.  Fig.  4,  to  represent  the  position  of  the  con- 
necting rod.  Let  the  line  a,  b,  represent,  by  its  length,  14,000  Ibs., 
which  is  the  tension  upon  the  rod.  Complete  the  rectangular  parallelo- 
gram a,  b,  c,  d,  then  will  the  line  a,  c,  represent  by  its  length  the 


upward  effect  of  the  tension  upon  the  connecting  rod.  Thus  the  length 
of  the  line  a,  b,  in  the  figure  is  6",  the  length  of  the  line  a,  c,  is  .5527". 
Therefore,  6  :  .5527  :  :  14,000  :  1.289.  Therefore  it  would  be  necessary 
to  exert  a  downward  force  of  1,289  Ibs.  to  keep  the  cross-head  from 
pressing  upon  the  upper  guide  at  this  point. 


Q.  35.  (1897-8.)  If  the  engine  of  Q.  28  has  a  cylinder  10"  internal 
diameter  and  is  running  with  a  gage  pressure  of  70  Ibs.,  what  is  the 
greatest  strain  brought  upon  the  connecting  rod?  The  weight  and  angu- 
larity of  the  connecting  rod,  lead,  back-pressure  and  compression  being 
neglected. 

Ans.  35.    We  have  the  force  of  inertia  and  the  steam  pressure  to 
take  into  account.     By  referring  to  Fig.  3  we  find  that  there  is  at  the 
commencement  of  the  stroke  a  .tension  of  about  18,000  Ibs.  on  the  con- 
ectmg  rod  due  to  the  inertia  of  the  parts;  but  the  action  of  the  steam 
to  produce  a  compression  upon  the  connecting  rod.     That  is   at  the 
commencement  of  the  stroke  the  steam  pressure  acts  in  the  opposite 
direction  to  the  force  of  inertia.     Therefore  the  strain  upon  the  con- 
necting rod  at  the  commencement  of  the  stroke  is  18000  —  70  X  7854 
(area  of  the  10"  piston)  =  18,000  —  5498  =  12,502  Ibs.  tension  upon  the 
connecting  rod.    At  the  end  of  the  stroke  the  force  of  inertia  is  15,310 

82 


2552  =  12,758  —a  force  of  compression;  to  this  should  be  added  the 

final  pressure  of  the  steam  upon  the  piston,  if  any.  Therefore,  the 
greatest  strain  brought  upon  the  connecting  rod  is  equal  to  about  12,758 
Ibs.  and  steam  pressure,  if  any,  and  is  a  force  of  compression  at  the 
end  of  the  stroke. 


Q.  36.  (1897-8.)  Draw,  to  scale,  a  diagram  of  the  engine  of  Q.  28 
that  shall  show  the  distances  traveled  by  the  piston  at  each  angular 
position  of  the  crank. 

Ans.  36.  With  the  compass  set  to  a  radius  equal  (to  a  convenient 
scale)  to  the  connecting  rod,  draw  the  circle  b,  c,  d,  i,  1,  Fig.  5.  Upon 
the  same  diameter  with  the  compass  set  to  a  radius  equal  to  the  length 
of  the  connecting  rod  plus  the  length  of  the  crank,  draw  a  circle  b,  e, 
f,  h,  k,  touching  the  first  circle  at  b.  Draw  radial  lines  a  e,  a  f,  a  h, 
from  the  center  a,  to  the  larger  circle.  Then  will  the  portion  of  said 
lines  between  the  two  circles  be  equal  to  the  travel  of  the  piston  at  the 
corresponding  angular  position  of  the  crank.  Thus,  when  the  crank  is 


in  the  position  a,  e,  the  travel  of  the  piston  is  equal  to  c,  e,  measured 
by  the  scale  adopted,  and  when  the  crank  is  in  the  position  a,  f,  then 
d,  f,  is  the  travel  of  the  piston  from  the  commencement  of  the  stroke. 

If  the  diagram  of  Ans.  36  is  drawn  around  the  same  center  as  the 
Zeuner  valve  diagram,  both  the  valve  and  piston  travel  will  be  shown 
for  each  angular  position  of  the  crank,  by  the  same  diagram. 

83 


Q  37  (1897-8.)  What  per  cent  of  the  steam  is  condensed  by  the 
cylinder  walls  in  simple  non-condensing,  fast  running  engines,  without 
steam  jackets,  of  from  50  to  100  HP.? 

Ans.  37.     From  25  to  50  per  cent. 

The  object  of  Q.  No.  37  is  to  call  attention  to  the  great  importance  of 
cylinder  condensation.  The  quantity  of  steam  condensed  being  much 
too  great  to  be  accounted  for  by  radiation  from  the  outside  of  the  cylin- 
der, and  only  to  be  accounted  for  upon  the  supposition  that  much  of  the 
heat  goes  to  re-evaporating  the  condensed  steam. 


Q.   41.      (1897-8.)      Does   the  fly-wheel   increase   the  power   of  the 
engine? 

Ans.  41.    It  regulates  but  does  not  increase  the  power. 


Q.  51.  (1897-8.)  What  should  be  the  shape  in  cross-section  of  the 
connecting-rod  of  a  fast  running  engine?  Why? 

Ans.  51.  The  cross-section  of  the  connecting  rod  in  fast-running 
engines  should  be  rectangular  and  greater  in  the  plane  of  motion  than 
perpendicular  to  said  plane.  In  his  paper  read  before  the  A.  S.  M.  E. 
Mr.  John  H.  Barr  finds  as  the  average  of  a  large  number  of  engines  that 
the  depth  is  2.7  times  the  breadth. 

The  depth  is  made  greater  in  order  to  resist  the  bending  action  of 
the  inertia  of  the  rod  itself. 


Q.  52.  (1897-8.)  Assuming  a  factor  of  safety  of  20  how  would  you 
determine  the  diameter  of  a  connecting  rod  having  a  circular  cross- 
section? 

Ans.  52.  To  calculate  the  diameter  of  a  circular  connecting-rod  may 
be  used  the  following:  — 

Rule: — Multiply  the  maximum  pressure  that  may  be  brought  upon 
the  rod,  in  pounds,  by  the  factor  of  safety,  and  extract  the  fourth  root 
(i.  e.,  the  square  root  of  the  square  root)  of  this  product.  Call  this 
result  No.  1. 

Extract  the  square  root  of  the  length  of  the  connecting-rod  expressed 
in  inches.  Call  this  result  No.  2. 

Multiply  result  No.  1  by  result  No.  2  and  divide  by  61.75.  This  will 
give  the  maximum  diameter  of  the  rod  in  inches. 

This  rule  is  expressed  algebraically  as  follows:  — 

1/PFVL 


61.75 

Suppose  we  take  the  following  data  for  example:  Area  of  piston 
100  sq.  ins.  Weight  of  reciprocating  parts  250  Ibs.  Speed  250  turns  per 
minute.  Maximum  steam  pressure  50  Ibs.  Stroke  1  ft.  Connecting-rod 
36"  long;  then  the  greatest  pressure  of  the  steam  on  the  piston  would  be 
100  X  50  =  5000  Ibs.  The  pressure  due  to  inertia  would  be  about 
250  X  13  X  13  -=-  16  =  2641  Ibs.  Therefore  the  maximum  pressure  to 
adapt  the  rod  to  is  5000  +  2641  =  7641  Ibs. 

Usirg  a  factor  of  safety  of  20,  we  would  have,  pressure  X  factor  of 
safety  =  7641  X  20  =  152,820.  The  square  root  of  152,820  is  about  391, 
and  the  square  root  of  391  is  about  19.77.  Therefore  the  fourth  root  of 
152,820  is  about  19.77.  This  is  result  No.  1. 

The  square  root  of  36,  the  length  of  the  connecting-rod    is  6. 

Six  times  19.77  is  118.62,  and  this  divided  by  61.75"  is  1.92,  say  a  2" 
Circular  connecting-rod. 

84 


In  Mr.  Barr's  paper,  above  referred  to,  the  average  factor  of  safety 
in  slow  running  engines  with  circular  rods  was  found  to  have  been  13. 

in  the  examples  of  Ans.  52  and  54,  the  pressure  due  to  the  inertia 
of  the  parts  is  taken  into  account.  In  some  cases  this  would  be  mate- 
rial, in  others  it  would  have  but  little  effect.  It  is  in  any  case  easily 
calculated. 

Cylindrical  and  square  parts  are  so  common  in  machine  construction 
that  the  rules  for  calculating  their  strength  are  very  useful.  Such  rules 
are  not  intended  to  give  exact  results,  they  are  intended  to  tell  us 
when  a  part  is  liable  to  be  overloaded,  and  when  it  is  safe.  For  this 
purpose  they  are  reliable. 


Q.  53.  (1897-8.)  How  are  the  dimensions  of  a  connecting-rod  having 
a  rectangular  cross-section  determined? 

Ans.  53.  The  method  of  securing  the  ends  of  the  connecting-rod 
would  require  that  the  height  (H)  of  the  cross-section  should  be  twice 
its  breadth  (B).  The  inertia  of  the  rod,  its  whipping  action,  would 
require  a  still  greater  relative  height  to  breadth. 

Mr.  Barr  found  the  minimum  H  to  be  2.2  B,  the  maximum  H,  4  B, 
and  the  average  H  to  be  2.7  B. 

Assuming  this  average  relative  value  of  H,  the  following  may  be 
taken  as  the 

RULE   FOR    CALCULATING   THE    DIMENSIONS    OF    A    RECTANGULAR   ROD. 

First.  Multiply  the  maximum  pressure,  in  pounas,  that  may  be 
brought  upon  the  connecting-rod,  by  the  factor  of  safety,  and  extract 
the  fourth  root  of  the  product.  That  is,  get  the  square  root  of  the 
square  root  of  this  product.  Call  this  result  No.  1. 

Second.  Find  the  square  root  of  the  length  of  the  connecting-rod 
(expressed  in  inches).  Call  this  result  No.  2. 

Third.  Multiply  result  No.  1  by  result  No.  2  and  divide  this  product 
by  127.8. 

This  will  give  the  breadth  of  the  rod  in  inches.  The  height  is  2.7 
times  the  breadth. 

This  is  expressed  algebraically  as  follows: 


127  8 

In  which,  as  before,  F  is  the  factor  of  safety,  P  the  maximum  pres- 
sure that  may  be  brought  upon  the  rod,  L  the  length  of  the  rod  in 
inches,  127.8  is  a  constant. 


Q.  54.  (1897-8.)  What  would  be  the  dimensions  of  a  proper  con- 
necting-rod for  the  engine  of  Q.  28  assuming  a  gage  pressure  of  70  Ibs.? 

Ans.  54.  The  pressure  due  to  the  inertia  of  the  parts  is  12758,  the 
maximum  pressure  due  to  the  steam  would  be  78.54  X  70  =  5498.  There- 
fore the  greatest  pressure  might  be  12758  -=-  5498  — 18256.  Assuming  a 
factor  of  safety  of  20  the  calculation  would  be  as  follows:  20  X  18256 
=  365,120.  The  square  root  of  365,120  is  about  604  and  the  square  root 
of  604  is  about  24.58.  Result  No.  1.  The  length  of  the  connecting-rod 
is  36";  the  square  root  of  this  6:  6  X  24.58  =147.48,  and  this  divided  by 
the  constant,  127.8,  gives  1.154  as  the  breadth  of  the  connecting-rod; 
2.7  X  1.154  =  3.1158  for  the  height  of  the  cross-section. 


Q.  57.  (1897-8.)  If  the  pinion  to  the  elevator  should  be  broken, 
what  data  would  you  send  to  the  factory  to  secure  a  wheel  to  replace 
it?  Give  specifications. 

Ans.  57.  Diameter  of  wheel,  diameter  of  shaft,  diametrical  pitch, 
width  of  face,  width  and  depth  of  keyway. 

85 


PREFACE  TO  1898-9  QUESTIONS  AND  ANSWERS. 
"FOUNDATIONS." 

The  utility  of  good,  firm  and  lasting  foundations  under  a  boiler, 
engine  or  other  moving  mechanism  about  a  steam  plant  will  hardly  be 
questioned,  and  while  some  situations  are  attended  with  natural  advan- 
tages favoring  this  end,  it  is  not  infrequently  the  case  that  considerable 
extra  labor  and  forethought  are  entailed  to  bring  about  the  desired 

Examples  of  poorly  constructed  and  overloaded  foundations  are 
abundant  and  their  general "  effect  too  well  known  to  require  much 
comment;  all  that  could  be  said  on  this  subject,  however,  might  be 
accepted  as  an  apt  simile  of  the  difficulties  encountered  by  engineers 
who  endeavor  to  build  '^'temples"  of  knowledge  upon  a  substructure 
inadequate  to  receive  it. 

The  absurdity  of  placing  a  hundred  horse-power  engine  on  a  ten 
horse-power  foundation  is  striking,  but  there  is  no  evident  reason  why 
this  is  not  to  be  considered  just  as  practicable  as  an  effort  to  acquire 
the  higher  branches  of  engineering  science  without  regard  to  the 
fundamental  principles  which  are  the  natural  approach  to  this  greatly 
coveted  goal. 

It  is  not  believed  that  this  slight  of  elementary  principles  is  wholly 
intentional  on  the  part  of  many  who  are  seeking  to  place  themselves  on 
a  higher  plane,  hence  if  a  few  pertinent  suggestions  will  serve  as  a 
beacon  to  some  of  our  workers  the  purpose  of  this  article  will  be  fully 
accomplished. 

Solutions  of  various  engineering  problems,  by  means  of  rules  and 
formulas,  do  not  afford  much  satisfaction  to  the  engineer,  who  is  not,  in 
a  general  way  at  least,  familiar  with  the  process  of  reasoning  upon 
which  the  final  result  depends.  By  a  rigid  application  of  the  several 
operations  indicated  by  the  mathematical  prescription  arranged  for  him, 
a  proper  answer  is  obtained,  but  the  "spice"  of  perfect  confidence  and 
assurance  which  should  be  felt  in  the  work  does  not  attend  the  result. 

To  safely  use  even  the  most  ordinary  engineering  formulae  it  is 
regarded  as  essential  that  a  keen  knowledge  of  ordinary  arithmetic  is 
at  ready  command,  and  further,  that  the  application  of  its  principles 
to  the  usual  methods  of  measuring  and  computing  surfaces  and 
volumes  has  been  mastered  well.  It  is  equally  important  that  a  precise 
knowledge  of  the  points  embraced  in  a  course  of  natural  philosophy 
should  be  well  founded  in  an  engineer's  mind,  and  while  engaged  in 
crowding  in  knowledge  at  the  "bung"  it  is  well  to  see  that  none  gets 
away  at  the  "spigot."  Should  the  conditions  just  noted  be  reversed,  it 
is  a  foregone  conclusion  that  the  case  is  just  about  as  hopeless  as  when 
the  "cask"  is  claimed  to  be  so  full  that  it  can  hold  no  more. 

Among  many  other  things  that  should  be  incorporated  in  an  engi- 
neer's "foundation"  of  knowledge,  no  other  is  of  more  value  than  abso- 
lutely correct  "notions"  concerning  the  methods  of  loading  a  structure 
and  thoroughly  understanding  the  strains  induced  thereby. 

Well  fixed  ideas  of  forces  and  their  different  effects  should  be  studied, 
for  it  is  the  effect  of  forces  that  the  engineer  has  to  do  with,  and  if  he 
fails  at  any  time  to  recognize  their  existence,  and  does  not  succeed  in 
bending  them  to  his  will  or  foretelling  their  possible  action,  it  is 
frequently  due  to  the  fact  that  he  may  not  know  enough  to  properly 
locate  them  and  predetermine  results  that  are  to  be  avoided. 

In  its  broadest  sense  engineering  stands  for  the  acquirement  of  the 
knowledge  necessary  to  determine  with  refined  accuracy  the  results  or 
possible  effects  due  to  the  action  of  forces;  it  makes  mankind  the  master 
over  matter  and  is  a  science  in  which  theory,  practice  and  reason  are 
combined  for  the  one  grand  purpose  of  furthering  the  ends  of  civiliza- 
tion. 

It  is  beyond  the  scope  of  this  article  to  do  more  than  merely  outline 
some  of  the  obstacles  that  seem  to  hinder  the  progress  of  our  earnest 

86 


plodders,  and  of  these  there  is  probably  no  one  other  hindrance  as 
great  as  the  common  fault  of  not  grasping  broadly  the  important  prin- 
ciples upon  which  the  whole  fabric  is  supported.  A  perfect  mastery  of 
rudiments  breeds  a  sort  of  intuition,  that  grows  unconsciously  and 
develops  the  man  into  the  successful  engineer. 

The  elementary  principles  of  mechanics  and  mathematics  should  be 
acquired  in  a  way  enabling  the  engineer  to  base  his  investigations  on  a 
knowledge  of  such  principles  as  are  actually  involved  in  any  given  case; 
he  must  be  able  to  detect  these  principles,  no  matter  how  "clothed"  or 
disguised,  and  his  reasoning  is  often  more  effective  for  a  given  purpose 
than  either  the  eye,  the  ear  or  the  other  senses — all  of  which,  however, 
must  be  more  or  less  acute  to  serve  his  ends. 

The  foundation,  therefore,  need  not  be  particularly  ostentatious; 
rather  see  that  the  several  courses  are  of  the  proper  material,  thor- 
oughly laid  and  well  cemented.  If  the  original  proportions  are  insuffi- 
cient to  carry  the  final  structure,  don't  continue  to  build,  but  tear  it 
down  to  the  bottom  and  start  again  with  a  broader  base  and  a  deeper 
footing.  C.  H.  F. 


INTRODUCTIONS  TO  QUESTIONS  43  TO  62   (1898-9). 

INTRODUCTORY. 

The  questions  under  the  heading  "Practical  Steam  Engineering," 
are  designed  to  provoke  general  discussion  at  association  meetings; 
the  aim  has  been  to  arrange  them  in  a  way,  tending  to  draw  out  indi 
vidual  opinions  without  the  need  of  special  preparation  on  the  part  of 
even  the  more  advanced  members.  No  special  pains  have  been  taken 
to  evade  points  upon  which  engineers  differ,  and  it  is  not  the  intention 
of  the  committee  to  allow  their  own  particular  preferences  to  interfere 
with  able  arguments  advanced  on  either  side  of  a  vexed  question. 

FIRST  LIST  OF  QUESTIONS. 

Note. — This  embraces  some  of  the  principal  practical  points  con- 
cerned in  the  installment  of  100  HP  engine  in  an  ordinary  manufactur- 
ing establishment.  Answers  based  on  the  assumptions  indicated  in  the 
first  question.  The  answers  selected  are  brief  and  to  the  point;  the 
information  conveyed  will  be  useful  to  all  engineers  who  may  be  new 
in  matters  pertaining  to  such  installations. 

The  committee  has  added  the  interlineations  which  are  set  out  in 
separate  paragraphs,  the  purpose  being  to  further  elucidate  the  topics, 
handled  in  the  answers  and  to  point  out  the  line  of  reasoning  followed 
in  arriving  at  the  results. 

The  comment  thus  offered  in  connection  with  the  answers  in  no  way 
detracts  from  the  worth  of  the  original  paper;  it  proves  the  figures  to 
be  correct  and  places  the  work  far  above  the  plane  of  a  good  guess. 
It  is  the  "reasoning"  that  represents  the  "kernel  in  the  nut." 


Q.  43.  (1898-9.)  A  simple,  non-condensing,  automatic  engine,  with 
releasing  valve-gear  is  to  be  chosen,  the  pressure  at  boiler  is  100  Ibs.  per 
square  inch  and  the  maximum  load  is  estimated  at  100  HP;  give  bore, 
stroke  and  speed  of  engine  you  would  select  to  develop  this  power 
economically. 

Ans.  43.  We  would  select  a  first-class  Corliss  engine  of  "world  wide" 
reputation — size  14"  X  36",  same  to  run  at  80  r.p.m. 

(Ed.  Com. — Majority  of  answers  favored  Corliss  engines  of  36" 
stroke.  Cylinder  diameters  in  fractional  inches  were  given  by  some 
and  though  such  sizes  would,  under  the  assumed  conditions,  develop 
the  required  100  HP  yet,  for  ordinary  purpose,  there  would  be  no 

87 


greater  need  for  a  IS1^"  dia.  steam  cylinder,  than  for  a  4*4"  dia.  pipe; 
commercial  sizes  should  be  adhered  to. 

Speeds  given  in  the  various:  answers  ranged  from  75  to  90  revolu 
tions. 

Answers  to  other  questions  are  based  on  engine  dimensions  given 
in  Ans.  No.  43.) 


Q.  44.  1898-9.)  Thirty-five  feet  of  overhead  pipe,  one  bend  and  two 
elbows  are  required  to  convey  steam  from  boiler  to  engine;  50  feet  of 
exhaust  pipe  extends  to  the  roof  and  discharges  to  atmosphere.  Give 
diameters  of  both  steam  and  exhaust  pipes;  state  if  steam  pipe  should 
drain  to,  or  from,  boiler  and  what  arrangements  are  necessary  for 
expansion  and  other  incidental  details,  also  how  feed  water  heater 
should  be  provided  for,  and  what  is  necessary  if  exhaust  steam  is  to  be 
utilized  in  heating  building. 

Ans.  44.     Diameter  of  steam  pipe  =  4". 

Diameter  of  exhaust  pipe  =  5". 

These  sizes  are  taken  from  well  known  formulas  for  steam  and 
exhaust  pipes  for  Corliss  engines  running  under  550  ft.  piston  speed 
per  minute. 

Steam  pipe  should  drain  toward  engine.  As  close  to  the  throttle 
valve  as  possible  should  be  placed  a  steam  separator  of  ample  capacity, 
correctly  drained  and  trapped,  and  of  approved  design.  Steam  pipe 
should  be  well  supported  and  so  arranged  as  to  not  tear  itself  to 
pieces  by  expansion  and  contraction.  We  prefer  all  bends  made  of 
pipe  bent  to  a  long  radius  for  first-class  work  in  lieu  of  cast  fittings  of 
any  description.  All  connections  should  be  flanged. 

Exhaust  pipe  should  be  led  to  a  heater  of  ample  capacity  (we  ap- 
prove of  an  open  heater  for  this  section)  and  pipe  properly  drained  in 
low  points  between  engine  and  heater.  The  exhaust  should  be  arranged 
with  a  proper  by-pass,  with  the  necessary  valves  so  that  heater  can  be 
shut  off  for  cleaning  and  repairs  during  operation  of  engine. 

For  exhaust  heating,  a  suitable  back  pressure  valve  with  ample 
area  should  be  provided. 

(Ed.  Com.— It  is  generally  conceded  that  velocity  of  steam  should  not 
exceed  6,000  ft.  per  minute  through  pipes  100  ft.  long.  The  flow  is 
always  assumed  to  continue  throughout  the  stroke,  and  with  this  as  a 
basis  we  may  analyze  the  given  case  as  follows: 

Engine  14"  X  36",  80  rev. — 480  ft.  piston  speed  per  min. 

Sq.  dia.  of  14   (cylinder)    196 

=  12.25. 

Sq.  dia.  of  4  (pipe)    16 

The  quotient  represents  the  ratio  of  cylinder  and  steam  pipe  areas. 
Piston  and  velocity  of  steam  will  be  in  like  proportion,  hence  piston 
speed  480  X  12.25  =  5880  ft.  which  is  "inside"  the  limit  prescribed. 

For  exhaust  pipes  a  velocity  of  4000  ft.  is  much  used;  a  comparison 
similar  to  the  above  gives  for  the  5"  pipe  specified  in  the  answer — 
196X480 

—  =  3763   feet. 
25 
Also  within  the  given  limit. 


Q.  45.  (1898-9.)  Give  initial  pressure  and  point  of  "cut  off"  assumed 
in  connection  with  your  answer  to  Q.  No.  43;  also  the  mean  effective 
pressure  corresponding  thereto  and  the  probable  weight  of  steam 
required  per  hour  for  each  horse-power. 

Ans.  45.  The  initial  gage  pressure  assumed  in  the  answer  to  Q.  No. 
•  Ibs.  per  sq.  in.  upon  piston. 

The  point  of  cut-off  for  100  HP  at  80  rev.  or  480  ft.  of  piston  speed 

.     88 


per  minute  will  equal  nearly  22%  of  the  stroke.  The  mean  effective 
pressure  will  equal  44  1/2  Ibs.  for  this  steam  pressure  and  point  of  cut- 
off. These  calculations  are  based  upon  an  assumed  clearance  of  4%  and 
an  absolute  back  pressure  of  16  Ibs. 

(Ed.  Com. — Figures  submitted  in  Ans.  No.  45  may  be  subjected  to 
the  following  line  of  reasoning: 

100   HP  =  3,300,000   ft.   Ibs. 

The  distance  or  space  is  represented  by  480  ft.  piston  travel  per  min., 
hence  3,300,000  -~  480  =  6875  is  the  corresponding  "weight  to  be  raised." 
This  "weight"  is  equally  distributed  on  the  face  of  a  14"  dia.  piston; 
hence  the  quotient  6875  divided  by  the  area  of  a  14"  circle  =  6875  -M53.9 
=  44.6,  which  is  the  mean  effective  pressure  necessary  for  a  100  HP 
under  the  specified  conditions,  and  which  agrees  substantially  with  the 
answer. 

Point  of  "cut-off"  given  at  22%  of  stroke  becomes  36  X. 22  =  7.92 
inches. 

The  clearance  at  4%  of  the  cylinder  volume,  that  is — 
153.9  X  36  X  .04 

Area  X  stroke  = —  =  1.44  in.  of  stroke. 

153.9 
Ratio  of  expansion. 

Stroke  +  clearance      36  +  1.44  =  37 . 44 

— —  A 

Cut-off  +  clearance  7.92  +  1.44=    9.36 
Terminal  pressure. 

Absolute  initial   90  +  14.7  =  104.7. 

=  26.17. 

Ratio  of  expansion  4 

Mean  pressure. 

Hyperbolic  log.  ratio  of  expansion  4  =  1.3863,  to  which  add  1,  giving 
2.3863. 

The  terminal  pressure  26.17  X  2.3863  =  62.43,  which  is  the  absolute 
mean  pressure  of  piston. 

The  answer  assumes  16  Ibs.  absolute  back  pressure;  hence  62.43  — 
16  =  44.43,  which  is  the  mean  effective  gage  pressure. 

Fixing  the  initial  pressure  at  90  Ibs.  with  100  Ibs.  at  boiler  provides 
for  "drop"  in  pressure  between  boiler  and  engine. 

Steam  required  per  horse-power  is  not  given.  It  is  safe  to  assume 
30  Ibs.  per  hour  for  average  conditions,  although  greater  economy  is 
possible  with  such  engines. 


Q.  46.  (1898-9.)  The  transmission  of  100  HP  from  engine  to  shaft 
is  by  belt;  explain  the  advantages  and  disadvantages  of  having  main 
driving  pulley  distinct  from  fly  wheel.  State  whether  this,  or  a  band 
wheel,  is  preferable,  provided  either  could  be  used. 

Ans.  46.  This  engine  is  to  transmit  its  power  by  belt,  consequently 
there  can  be  no  valid  reason  for  the  use  of  a  separate  fly  wheel.  There 
may  be  cases  where  it  is  desirable  to  use  a  separate  band  wheel,  but 
this  being  a  new  installation  it  should  be  arranged  without  it.  Its 
disadvantages  in  this  particular  case  would  be  extra  cost  and  a  display 
of  poor  engineering  judgment. 


Q.  47.  (1898-9.)  The  full  power  of  the  100  HP  engine  referred  to, 
in  Question  No.  43,  is  to  be  belted  to  a  shaft  which  is  to  run  at  240  revo- 
lutions per  minute.  Give  diameters,  weight  and  face  of  band  wheel 
you  would  approve  of;  diameter  of  first  driver  pulley  and  best  and 
most  convenient  position  of  engine  with  reference  to  shaft. 

Ans.  47."  A  band  fly  wheel  for  this  engine  should  be  for  80  rev.  per 
min.,  10  ft.  diam.,  with  a  weight  of  9,000  Ibs.  Face  should  be  20"; 

89 


engine  running  80  rev.,  band  wheel  10  ft.  dia.,  to  drive  first  driven  pulley 
at  240  rev.  we  have  10  ft.  =  120  ins.;  hence 
80  X  120 

240 


.  fo'mul'a  for  "fly  whee,.  Cor.iss  eng.nes 

running  at  48fr  ft.  piston  speed,  is  as  follows: 

W  =  700,000  X =  7717.5 

D2  X  R2 

W  =  weight  of  wheel  in  pounds, 
d  =  dia.  of  cylinder  in  inches. 
W  s  =  stroke  in  inches. 

D  =  dia.  of  wheel  in  feet. 
R  =  rev.  per  min. 

According  to  Thurston's  formula  the  constant  number  700,000  be- 
comes 785  400   and  applied  in  that  form  to  the  case  in  hand  we  have— 
196  X  36 

W  =  785,400  X =  8659  lbs.« 

100  X  6400 

The  answer  agreeing  substantially  with  the  round  number  given  in 
repiy  to  the  question. 


Q.  48.  (1898-9.)  What  width,  double  leather  belt,  should  be  pro- 
vided to  transmit  satisfactorily  100  HP  according  t:>  the  condition  not- 
ed in  Q.  47?  State  what  you  consider  the  best  method  of  joining  the 
belt  and  at  what  angle  should  same  be  run  for  l/est  results? 

Ans.  48.  The  belt  should  be  oak-tanned  stoc'j,  short  lapped  and 
double;  width  should  be  18  ins. 

The  best  and  only  Al  method  of  joining  a  belt  of  this  size  is  a  care- 
fully lapped  and  cemented  joint. 

For  best  conditions  the  belt  should  not  be  lea  off  from  the  main 
shaft  at  a  greater  angle  than  45°  from  the  horizontal.  *  *  * 

(Ed.  Com. — When  the  size  of  a  oelt  is  determined  by  one  of  the 
numerous  rules  on  the  subject,  it  is  well  to  size  up  the  result  from 
practical  points  of  view  before  finally  deciding  the  question.  Liberality 
in  belt  proportion  is  a  good  investment.  The  need  of  extremely  tight 
belts  should  be  avoided. 

According  to  some  formulas  on  the  subject,  an  18"  double  belt  is 
quite  small  for  100  HP;  by  another  rule  it  is  ample,  and  to  choose  safely 
between  such  extremes  is  a  matter  of  judgment. 


Q.  49.  (1898-9.)  How  wide  and  how  deep  should  the  trench  for  an 
engine  foundation  be  excavated,  in  solid  earth,  and  what  modifications 
would  be  deemed  necessary  if  the  conditions  were  less  favorable? 

Ans.  49.  The  length,  depth  and  width  of  trench  for  this  engine 
foundation  would  depend  upon  the  character  of  the  soil. 

The  length  for  good  conditions  would  be  approximately  24  ft.  Depth, 
8  ft,  with  12"  of  concrete  before  beginning  brick  work.  The  widtn 
opposite  main  bearing  for  outboard  bearing  will  depend  upon  the  dis- 
tance between  main  bearing  and  center  of  outboard  bearing,  as  varying 
lengths  of  shafts  require  different  measurements  between  bearings,  and 
consequently  different  brickwork  as  to  width  at  this  place.  The  width 
of  the  "L"  part  of  this  trench  should  approximate  13  ft. 

In  many  places  it  is  found  desirable  to  drive  piling  and  upon  that 
place  a  proper  grillage  before  ramming  in  concrete.  This  will  satisfy 
all  but  extraordinary  conditions. 

90 


Q.  50.  (1898-9.)  Which  is  the  preferable  material  for  an  engine 
foundation — brick  or  stone,  and  how  should  same  be  laid? 

Ans.  50.  The  best  material  for  engine  foundations,  all  things  con- 
sidered, is  good  hard-burned  brick  laid  in  cement  mortar  and  every 
course  wet  down  and  grouted. 


Q.  51.  (1898-9.)  How  should  the  fly  wheel  pit  be  proportioned  and 
built? 

Ans.  51.  The  pit  should  be  proportioned  for  a  reasonable  amount  of 
room.  The  walls  should  be  washed  and  plastered  with  first-class  ce- 
ment. Also  the  bottom  should  be  treated  likewise,  and  should,  if  possi- 
ble, be  connected  to  the  sewer  or  its  equivalent. 


Q.  52.  (1898-9.)  What  precautions  are  to  be  observed  in  making, 
handling  and  setting  a  foundation  template? 

Ans.  52.  In  a  foundation  template  the  center  lines  of  cylinder  and 
main  shaft  should  be  square  with  each  other  and  all  bolt  holes  very 
carefully  located  as  per  foundation  drawing.  It  should  be  well  braced 
to  preserve  its  alignment.  It  should  be  handled  with  care.  It  should 
be  set  at  the  proper  height  and  carefully  leveled,  and  also  lined  to  exist- 
ing lines,  if  any  have  been  run,  and  properly  made  secure  to  preserve 
its  correct  position. 


Q.  53.  (1898-9.)  Explain  how  foundation  bolts  should  be  set  and 
walled  up.  Describe  also  the  form  of  fastening  preferred  for  lower 
ends  of  same. 

Ans.  53.  Foundation  bolts  should  be  placed  in  template  holes  for 
their  reception,  the  tops  raised  to  their  proper  heights  and  the  bolts 
plumbed  vertically.  Timbers  should  be  slid  over  the  bolts  with  room 
enough  for  plenty  of  lateral  play.  The  lower  ends  are  best  made  secure 
by  nuts  placed  in  "pocket  plates."  Recesses  should  be  placed  in  bottom 
of  foundation  so  that  these  can  always  be  reached.  The  bottom  threads 
should  be  long  and  the  point  of  bolt  well  leveled,  so  in  case  of  necessity 
they  can  easily  be  screwed  in  from  top  of  foundation. 


Q.  54.  (1898-9.)  Would  you  use  artificial  or  natural  stone  for 
coping  or  capstone?  What  advantage  has  one  over  the  other  in  engine 
foundations? 

Ans.  54.  There  is  but  very  little  advantage  of  one  over  the  other, 
providing  the  artificial  stone  is  properly  made.  It  usually  resolves  itself 
into  a  question  of  cost  and  facilities  at  hand  for  procuring  a  natural 
stone.  Whichever  are  used,  they  should  be  always  painted  to  exclude 
any  oil. 


Q.  55.  (1898-9.)  Which  are  the  best  for  leveling  up  preparatory  to 
bedding — iron  or  wood  wedges?  What  precautions  are  to  be  observed 
in  this  operation? 

Ans.  55.  Iron  wedges  are  better  than  wooden  ones  for  this  work. 
It  depends  much  more  upon  the  ability  of  the  erecting  engineer  than 
upon  the  material  that  the  wedges  are  made  of.  Many  men,  erecting 
engines,  lack  judgment  in  driving  wedges  of  all  kinds. 

In  driving  wedges  it  is  necessary  that  they  be  of  the  same  uniform 
taper;  that  they  are  placed  in  correct  positions  under  the  several  parts 
of  the  frame,  and  that  they  be  most  carefully  driven  so  as  to  produce 
no  distortion. 


Q  56  (1898-9.)  Explain  how  to  bed  the  100  HP  engine  assumed  m 
Q  No  43  noting  the  materials  you  believe  best  adapted  for  the  purpose. 

Ans  56  The  engine  must  be  placed  upon  the  foundation,  carefully 
lined  up  and  leveled  up  on  wedges.  A  space  of  about  5/16"  or  so  should 
be  left  between  foundation  stones  and  bottom  of  frame.  This  space 
should  be  filled  in  solidly  with  a  thin  grouting  of  2  parts  best  Portland 
cement  to  1  part  of  sharp  sand.  If  the  engine  has  a  girder  bed  with  a 
center  bearing  under  front  end  of  girds,  this  should  be  left  until  the  last 
before  grouting.  After  joint  has  hardened  thoroughly  the  wedges 
should  be  backed  out,  all  joints  pointed  and  trimmed  and  the  joint 
well  painted  to  exclude  all  oil. 


Q.  57.  (1898-9.)  How  would  you  treat  the  foundation  and  the  set- 
ting of  the  outboard  bearing? 

Ans.  57.  The  outboard  bearing  should  receive  the  same  careful 
attention  both  as  to  material  and  care  in  bedding  and  should  be  square 
with  center  line  of  engine  and  lie  in  the  same  horizontal  plane. 

Before  starting  the  shaft  should  be  rolled  over  in  its  bearings  by 
hand  and  then  raised  and  the  babbit  linings  scraped  to  a  good  fair 
bearing. 

Q.  58.  (1898-9.)  What  two  tests  can  readily  be  applied  to  an  ordi- 
nary spirit  level  to  prove  its  fitness  for  use  in  lining  up  an  engine? 

Ans.  58.  A  spirit  level  can  be  tested  for  truth  by  the  usual  method 
of  reversing  ends  and  noting  position  of  bubble  or  by  the  rather  unusual 
method  of  an  accurate  square  and  plumb  bob. 

The  following  from  "Instruction  to  Engineers,"  issued  by  the  Buck- 
eye Engine  Company,  is  of  interest  in  this  connection: 

"Good  spirit  levels  will  be  needed  for  this  work — better  ones  than 
can  usually  be  had  at  the  stores — one  iron-bodied  one  in  particular,  3" 
to  6"  long,  to  be  applied  crosswise  on  slides  and  elsewhere. 

"The  test  of  a  good  level  is  the  distance  its  bubble  travels  for  a  given 
amount  of  departure  from  the  true  level.  To  test  a  level  place  it  on  a 
planed  strip  of  board  or  plank  two  feet  long  (if  the  level  itself  is  not  a 
two-foot  wooden  one) ;  support  the  ends  of  the  strip  on  blocks  close  to 
ends,  so  adjusted  that  the  bubble  will  be  in  center.  Then  lift  one  end 
and  insert  a  piece  of  postal  card  or  other  card  of  about  the  same  thick- 
ness (a  postal  card  is  full  1-100"  thick)  and  note  how  much  the  bubble 
has  moved.  It  should  have  moved  quite  perceptibly;  if  about  1/8", 
and  same  whichever  end  is  raised,  it  is  very  good,  if  1/16",  fairly  good, 
but  requires  very  close  observation." 


Q.  59.  (1898-9.)  What  methods  or  devices  would  you  use  in  lining 
up  the  crank-shaft? 

Ans.  59.  For  this  work  we  would  use  a  fine  waterproof  silk  line  for 
drawing  through  center  of  cylinder  and  girder  and  extend  it  beyond 
main  bearing  a  convenient  distance.  Bring  the  crank-pin  up  to  the  line 
in  nearly  extreme  forward  and  backward  positions;  when  the  distance 
from  line  to  face  of  crank-pin  hub  measures  alike  when  crank-pin  is 
brought  up  to  the  line  in  the  two  different  positions,  the  shaft  is  square 
laterally.  A  spirit  level  can  be  used  to  adjust  shaft  in  the  other  direc- 
tion to  the  best  advantage  in  this  particular  case. 


Q.  60.  (1898-9.)  Outline  briefly  the  successive  steps  and  precautions 
to  be  observed  before  a  newly-placed  engine  is  ready  for  steam? 

Ans.  60.  Before  connecting  to  engine  permanently,  a  new  steam  pipe 
should  be  thoroughly  blown  out  with  steam  of  good  pressure.  The  steam 
chamber,  valve  ports,  valves,  piston,  etc.,  should  be  carefully  cleaned  of 
all  foreign  matter  and  liberally  supplied  with  good  cylinder  oil;  also  a 

92 


generous  application  of  flake  graphite.  The  striking  points  of  piston 
and  also  clearance  of  same  should  be  determined  and  properly  marked. 
Valve  stems  and  piston  should  be  properly  packed.  The  valves  should 
be  properly  set.  Everything  should  be  made  secure  and  the  bearings 
properly  adjusted.  The  engine  should  be  turned  over  by  hand  one 
complete  revolution  and  moving  parts  carefully  watcned  for  interference 
and  fouling,  one  part  with  another.  Gradually  warm  up  cylinder  and 
start  cylinder  lubricator  and  adjust  all  oil  cups  for  a  liberal  supply  of 
oil  at  first. 

After  engine  uas  run  long  enough,  say  three  or  four  days,  always 
apply  indicator  and  make  final  adjustments  to  valves. 


Q.  61.  (1898-9.)  Indicator  cards  taken  prove  the  engine  to  under- 
loaded and  likely  to  remain  so  for  some  time  to  come — what  course 
would  you  pursue? 

Ans.  61.  First  reduce  steam  pressure.  Second,  reduce  speed  if  the 
character  of  the  load  as  regards  regulation  will  admit  of  it.  Third, 
in  extreme  cases  it  may  be  necessary  to  reduce  both. 


Q.  62.  (1898-9.)  Give  an  approximate,  itemized  estimate,  covering 
the  cost  of  the  outfit  outlined  in  the  preceding  questions,  freights,  cart- 
age or  extraordinary  conditions  not  to  be  considered. 

Ans.  62.  The  following  estimate  is  based  upon  Al  material  and 
work: 

14"  X  36"  Corliss  engine  complete $1,600 

Foundation  complete,  cap  and  bottom  stones 450 

Piping  complete,  including  separator,  4" 200 


Total   $2,250 

(The  lowest  estimate  received  was  $1,762.    The  highest  was  $3,539.75, 

and  includes  an  indicator  outfit  and  all  incidentals.     The  average  of 

all  estimates  is  $2,233. 

The  asking  for  estimates  seemed  an  innovation,  yet  it  is  proper  that 

men  should  have  some  sort  of  a  notion  of  the  actual  money  value  of  the 

apparatus  in  their  care.    A  discussion  of  the  "cost  of  things"  is  a  topic 

worthy  of  consideration. — Ed.  Com.) 


Q.  67.  (1898-9.)  Steam  pressure  at  boilers  is  100  pounds;  give 
bore,  stroke  and  revolutions  per  minute  for  a  100  HP  high-speed,  simple 
and  direct  connected  automatic  engine,  suitable  for  work  of  the  char- 
acter indicated  in  Q.  65;  the  power  to  be  developed  within  the  economi- 
cal range  of  the  engine,  all  general  conditions  assumed  to  be  favorable. 

Ans.  67.  For  100  Ibs.  steam  pressure  at  boilers  the  following  pro- 
portions are  deemed  suitable  for  100  HP  engine  of  the  "high-speed" 
type  referred  to  in  the  question. 

Diameter  of  cylinder,  12";  stroke  of  piston,  12".  Revolutions  per 
minute,  300.  Piston  travels  600  ft.  per  minute. 

Average  mean  effective  pressure  for  100  indicated  HP  =  49  Ibs. 


Q.  68.  (1898-9.)  Why  are  "high-speed"  engines  of  the  type  referred 
to  in  Q.  67  considered  well  adapted  for  electrical  installations  of  the 
size  noted  in  the  preceding  question? 

Ans.  68.  High  speed  engines  of  the  type  referred  to  in  Q.  67  require 
less  floor  space  per  horse-power  and  are  better  adapted  for  direct  con- 
nection to  dynamo;  they  afford  better  regulation  and,  by  a  higher  speed 
of  rotation,  effect  a  saving  and  a  resultant  smaller  size. 


Q.  69.  (1898-9.)  What  are  the  relative  advantages  of  horizontal 
and  vertical  engines  and  what  reasons  govern  your  selection,  assuming 
"make-up"  of  both  types  to  be  about  the  same? 

93 


Ans  69  Assuming  the  "make-up"  of  horizontal  and  vertical  engines 
to  be  about  the  same  the  only  advantage  the  former  have  over  the  other 
is  in  the  matter  of  head  room  or  height. 

In  favor  of  the  vertical  engine  may  be  noted:  Economy  of  floor 
space  and  the  fact  that  piston  and  valves  are  less  affected  by  friction 
and  unequal  wear  and  therefore  require  less  lubrication. 

It  is  largely  a  question  of  location  and  the  ground  room  at  com- 
mand- better  accessibility  of  the  working  parts  is  a  point  argued  in 
favor 'of  the  horizontal  engine.  For  very  large  engines  the  tendency 
of  modern  practice  is  decidedly  towards  vertical  engines. 

070  (1898-9.)  Submit  sketch,  showing  by  indicator  card,  the  con- 
ditions you  expect  to  realize  in  the  cylinder  of  the  engine  selected  when 
driving  the  normal  load  indicated  by  Q.  67. 

Ans.  70.    See  drawing  herewith: 

—  Card  JVo.J  - 

•Sea  Ze  of  Sp-ri77<? 


ANSWER  No.  70. 

Q.  71.  (1898-9.)  Under  the  conditions  noted  in  Q.  70,  what  will  be 
the  probable  steam  consumption  in  pounds,  per  hour,  for  each  horse- 
power? State  also  how  this  compares  in  economy  with  the  ordinary, 
well-made,  throttling  slide-valve  engine,  performing  a  like  duty? 

Ans.  71.  Under  conditions  noted  in  question  70  the  best  type  of 
single-valve,  simple,  non-condensing  engine,  should  produce  an  indicated 
horse-power  on  32  Ibs.  of  steam  per  hour. 

The  excess  above  figure  quoted,  depends  on  losses  due  to  conditions 
of  valves,  etc. 

For  throttle  governed  slide  valve  engine  cutting  off  5/8  stroke,  40 
Ibs.  is  usually  allotted,  although  a  less  amount  is  claimed  for  such 
engines  when  in  good  condition. 


Q.  72.  (1898-9.)  Has  a  "cross-compound,"  slide-valve  engine,  fitted 
with  throttling  governor,  any  special  advantages  for  the  work  outlined 
in  the  preceding  questions? 

Ans.  72.  There  are  many  advantages  of  a  cross-compound,  throttling 
engine  over  a  single  valve,  automatic  engine.  They  are  very  simple, 

94 


show  a  good  economy  of  steam  and  maintain  this  economy  over  a  wide 
range  of  power  and  can  be  kept  free  from  leakage. 

(Ed.  Com. — The  foregoing  is  one  response  favoring  the  type  of 
engine  referred  to.  Some  correspondents  would  not  allow  any  advan- 
tages— answering  simply  "none."  For  the  work  outlined  in  the  ques- 
tion the  throttle  governed  cross-compound  would  probably  be  equally  as 
economical;  it  is  questionable,  however,  if  regulation  would  be  as  good 
as  in  a  first-class  automatic  engine.) 


Q.  6^..  (1899-1900.)  If  a  steam  jacket  is  used,  is  the  steam  in  the 
cylinder  affected  by  the  heat  of  the  steam  in  the  jacket?  If  so,  explain 
how. 

Ans.  64.  It  is  assumed  that  the  steam  in  the  cylinder,  while  ex- 
panding, receives  just  enough  heat  from  the  steam  in  the  jacket  to 
prevent  any  appreciable  part  of  it  from  condensing,  without  super- 
heating the  steam  in  the  cylinder,  reducing  the  condensation  in  the 
cylinder  to  a  minimum,  the  maximum  condensation  taking  place  in 
the  jacket.  In  reference  to  advantages  and  economy,  see  "Kent,"  page 
787,  and  other  noted  authors  on  the  subject. 


Q.  66.  (1899-1900.)  Is  there  any  gain  in  using  steam  at  100  Ibs. 
gauge  pressure,  expansively,  with  a  mean  effective  pressure  (M.E.P.) 
of  70  Ibs.,  over  using  steam  at  70  Ibs.  the  whole  length  of  the  cylinder? 
For  illustration,  assume  a  cylinder  with  a  volume  of  3  cu.  ft. 

Ans.  66.  Yes,  there  is  a  gain  in  using  steam  expansively  in  the 
conditions  as  stated  by  the  question. 

Using  a  cylinder  with  a  volume  of  3  cu.  ft.  and  an  initial  pressure 
of  70  Ibs.,  continued  throughout  the  full  stroke,  would  be  using  3  cu.  ft. 
of  steam  at  70  Ibs.  pressure,  or  a  weight  of  .603  Ibs.  of  steam. 

If  the  initial  pressure  is  raised  to  100  Ibs.,  and  with  a  M.E.P.  of 
70  Ibs.,  the  cut-off  would  take  place  at  %  of  the  stroke;  therefore  at 
this  pressure  there  would  be  %  of  3  cu.  ft.  of  steam  used  to  fill  the 
required  volume  of  the  cylinder. 

The  weight  of  steam  at  100  Ibs.  pressure  equals  .264  Ib.  per  cu.  ft. 

%  of  .264  Ibs.  =  .099  Ib.  used  for  1  cu.  ft.  of  the  cylinder;  .099  X  3 
=  .297  Ib.  =  the  proportional  quantity  of  steam  used  in  a  cylinder  with 
a  volume  of  3  cu.  ft.  Therefore,  .603  Ib.  steam  used  at  70  Ibs.  pressure 
—  .297  Ib.  steam  used  at  100  Ibs.  pressure  =  the  amount  of  steam 
gained  in  weight  by  working  the  steam  expansively  at  each  stroke,  or 
one-half  revolution. 

.603  —  .297  =  .306  Ib.,  an  approximate  gain  of  51  %,  .306  H-  .603. 

Second  answer  (using  the  same  conditions). 

Taking  a  cylinder  of  a  volume  of  3  cu.  ft.,  we  will  assume  a  piston 
area  of  1  sq.  ft.,  or  144  sq.  in.  Discarding  clearance,  we  will  only 
consider  as  a  back  pressure  that  of  the  atmosphere — viz.,  14.7  Ibs.  per 
sq.  in. — and  using  the  steam  expansively  with  the  initial  pressure  of 
100  Ibs.;  70  Ibs.  M.E.P. 

One-third  cut-off,  theoretically,  gives  for  100  Ibs.  g.p.  69.96  Ibs. 
M.E.P.;  practically,  70  Ibs. 

In  practice  it  is  customary  to  deduct  at  least  4  Ibs.  from  the  theo- 
retical M.E.P.,  making  it  to  conform  to  that  determined  from  actual 
practice;  making  the  cut-off  %  of  the  stroke,  gives,  theoretically,  74 
Ibs.  M.E.P.;  deducting  4  Ibs.  gives  us  the  70  Ibs.  M.E.P.  required. 

Hence  we  have — 

Stroke  =  3  ft.;   144  sq.  in.  piston  area. 

Cut-off  =  3/8  stroke  =  13i/2  in.,  or  1.125  ft. 

Initial  pressure  =  100  Ibs.  g.p. 

Back  pressure  =  14  7-10  Ibs.,  or  atmosphere. 

Then  3' 4-1.125'  =  2.4  =  ratio  of  expansion. 

Hgp.  Log.  +  1  of  the  ratio  of  expansion  =  1.8755;  then 

95 


1.8755X100X144  =  27007.2   foot  pounds,   the  total  work   done  on 

The  negTuve^work,   or  back   pressure,   expressed   ir    foot   pounds, 
equals 


on  the  pst™  at  100  Ibs.  g.p.  used  expansively,  with  70  Ibs.  M.B.P.  for 

°nV*Wng  the  other  condition,  using  steam  at  70  Ibs.  g.p.,  the  whole 
length  of  cylinder,  at  a  constant  pressure,  we  have 

144  X  70  X  3  =  30240  foot  pounds,  as  to  the  total  work  done  on  the 

PiSAs'the  negative  work  will  be  the  same  as  in  the  preceding  condi- 
tl0?0240  —  6350.4  =  23889.6  foot  pounds  of  work  done  on  the  piston 

%h7d7fference  between  the  two  conditions  will  be  23889.6-20666.8 
=  3232  8  foot  pounds  more  work  done  at  a  constant  pressure  of  70  Ibs. 
than  when  using  the  steam  expansively  at  70  M.E.P. 

But  in  using  the  steam  at  a  constant  pressure  of  70  Ibs.  we  have 
used  3  cu.  ft.  of  steam,  at  a  weight  of  .1971  Ib.  per  cu.  ft;  .1971  X  3  = 
.5913  Ib.  steam  per  one  stroke. 
At  100  Ibs.  g.p.,  %  cut-off. 
The  steam  weighs  .2617  Ib.  per  cu.  ft.;  hence 
.2617  X  1.125  =  .2944  Ib.  steam  used  expansively. 
Therefore, 

5913 2944  =  .2969  Ib.  of  steam  used  at  a  pressure  of  70  Ibs.  con- 
stant, doing  3232.8  foot  pounds  more  work  than  when  used  expansively 
at  100  Ibs.  initial  pressure  proportionally. 

The  amount  of  steam  used  in  the  two  conditions  = 
.5913  :  .2944  :   :  100  :  x. 

x  =  49  8-10%. 

The  percentage  gained  in  using  the  steam  expansively  = 
.5913  :  .2969  :   :  100  :  x. 
x  =  50  2-10  %  gained. 

Note — Should  this  example  be  worked  out,  on  the  theoretical  basis 
of  1/3  cut-off,  the  percentage  gained  =  55  7-10%. 


Q.  87.  (1899-1900.)  What  will  be  the  centrifugal  force  developed  by 
a  fly-wheel  eighteen  feet  in  diameter,  thickness  of  rim  five  inches, 
width  of  face  forty-three  inches,  turning  at  a  rate  of  90  R.P.M.  (the 
arms  and  hub  not  taken  into  account)  ?  The  weight  of  a  cubic  inch 
of  cast  iron  taken  as  .261  pounds. 

Ans.  87.  Centrifugal  Force:— When  a  body  moves  in  a  circular 
path  it  tends  at  each  point  to  move  in  a  tangent  to  the  circle  at  that 
point.  But  at  each  point  it  is  deflected  from  the  tangent  by  a  force 
acting  toward  the  center  of  the  circle.  This  force  may  be  the  tension 
of  a  string  or  the  arm  of  a  fly-wheel,  or  the  attraction  between  a  planet 
and  its  moon,  or  the  inward  pressure  of  the  rails  on  a  curve,  etc. 

Like  all  force,  it  is  an  action  between  two  bodies,  tending  either 
to  separate  them  or  to  draw  them  closer  together,  and  acting  equally 
upon  both.  In  the  case  of  a  string  it  pulls  the  body  towards  the  center, 
and  the  hand  towards  the  body,  i.  e.,  from  the  center.  In  the  case  of 
the  car  on  a  curve  it  pushes  the  car  toward  the  center  and  the  rails 
from  the  center. 

The  push  or  pull  on  the  revolving  body  toward  the  center  is  called 
the  centripetal  force;  while  the  pull  or  push  tending  to  move  the 
deflecting  body  from  the  center  is  called  the  centrifugal  force. 


These  two  forces,  being  merely  the  two  "sides"  (as  it  were)  of  the 
same  stress,  are  necessarily  equal,  and  opposite,  and  can  only  exist 
together. 

The  moment  the  stress  or  tension  exceeds  the  strength  (or  inherent 
cohesive  force)  of  the  string,  or  substance  the  latter  breaks.  The 
centripetal  and  centrifugal  forces  therefore  instantly  cease;  and  the 
body  no  longer  disturbed  by  a  deflecting  force,  moves  on,  at  a  uniform 
velocity  in  a  tangent  to  its  circular  path,  or  at  right  angles  to  the 
direction  which  the  centrifugal  force  had  at  the  moment  it  ceased. 

The  total  centrifugal  force  exerted  or  developed  in  a  fly-wheel  18 
ft.  diameter,  43  in.  face,  5  in.  thickness  of  rim,  is  found  by  the  follow- 
ing rule: 

Rule:  —  Multiply  the  weight  of  the  rim  in  pounds  by  the  velocity  in 
feet  per  second,  squared;  divide  this  product  by  tl  3  value  of  "g"  (32.2') 
multiplied  by  the  mean  radius  in  feet. 

Expressed  in  formula: 


gR 

Let  F  =  total  cent,  force. 
Let  W  =  weight  of  rim  in  Ibs. 
Let  R  =  mean  rad.  in  feet. 
Let  v  =  veloc.  in  ft.  per  sec. 
Let  g  =  rate  of  accel.  in  ft.  per  sec. 
Let  r  =  R.  P.  M. 
Let  D  =  mean.   dia.   in  inches. 
Let  v  =  3.1416. 

W  =  Dir  (43  X  5  X.261) 
W  =  211  X  3.1416  X  43  X  5  X  .261  =  37197.5   Ibs. 

R  —  9'  _  2W  =  8.79' 
2,rRr 

60 

2  X  3.1416  X  8.79  X  90 
V  =  -  —  =  82.844  ft. 

60 
According  to  our  formula  — 

Wv2 

F  =  - 
gR 

37197.5  X  S2.8442       37197.5  X  6863.13 
F=—  —  =  — 

32.2  X  8.79  32.2  X  8.79 

255291278.175 

F  =  -  =  901968.     pounds. 
283.038 

F  =  901968   Ibs..  the  total   centrifugal   force   developed   by  the  fly- 
wheel. 

By  another  formula  —  using  same  value  of  letters— 

4  WWR  r2 
F  =  -    —  —  =  902000  Ibs. 

3600  g 

A  variation  of  only  32  Ibs.  between  the  results  of  two  formulas, 
probably  caused  in  disuse  of  decimals. 

By  another  formula  using  same  value  of  letters  — 

F  =  .000341  W  R  r2  =  902016  Ibs. 
A  variation  of  only  16  Ibs.  from  second  formula. 
A  variation  of  48  Ibs.  from  first  formula. 

97 


o  88  (1899-1900.)  What  factors  enter  into  the  transmitting  capac- 
ity of  leather  belts?  How  is  the  transmitting  capacity  calculated? 

Ans  88.  Belts  and  Belting:— Although  there  is  not  nearly  as  much 
known  in  general  about  the  power  of  transmitting  agencies  as  there 
should  be,  still  it  seems  that  almost  any  other  method  or  means  is 
better  understood  than  belts. 

One  of  the  chief  difficulties  in  the  way  of  a  better  understanding,  or 
knowledge,  of  the  belting  problem  is  the  relation  that  the  belts  and 
'pulleys  bear  to  each  other.  The  general  supposition,  and  one  that 
leads  to  many  errors,  is  that  the  larger  in  diameter  a  pulley  is  the 
greater  its  holding  capacity— the  belt  will  not  slip  so  easily  is  the  belief. 
But  it  is  merely  a  belief,  and  has  nothing  to  sustain  it  unless  it  be 
faith,  and  faith  without  works  is  an  uncertain  factor.  The  following 
immutable  principles,  or  laws,  should  be  impressed  upon  the  minds  of 
all  interested  in  this  subject: 

First — The  adhesion  of  the  belt  to  the  pulley  is  the  same — the  arc 
or  the  number  of  degrees  of  contact,  aggregate  tension  or  weight  of 
the  belt  being  the  same — without  reference  to  the  width  of  the  belt  or 
diameter  of  the  pulley. 

Second — A  belt  will  slip  just  as  readily  on  a  pulley  4  feet  in  diam- 
eter as  it  will  on  a  pulley  2  feet  in  diameter,  provided  the  condition  of 
the  faces  of  the  pulleys,  the  arc  of  contact,  the  tension  and  the  number 
of  feet  the  belt  travels  per  minute,  are  the  same  in  both  cases. 

Third — A  belt  of  a  given  width,  making  two  thousand  or  any  other 
given  number  of  feet  per  minute,  will  transmit  as  much  power  running 
on  pulleys  two  feet  in  diameter  as  it  will  on  pulleys  four  feet  in  diam- 
eter, provided  the  arc  of  contact,  tension,  speed  and  condition  of  pulley 
faces  all  be  the  same  in  both  cases,  hence — 

The  principal  factors  entering  into  the  transmitting  capacity  of  belts 
are  the  speed  in  feet  per  minute,  arc  of  contact,  the  position  in  which 
it  travels,  the  thickness,  weight  and  the  condition  of  the  belt,  to  adhere 
to  the  face  of  the  pulley. 

The  transmitting  capacity  of  belts  is  calculated  by  the  following 
rule: 

Allowing  1  inch  in  width  traveling  at  a  given  speed  in  feet  per 
minute  to  transmit  a  HP,  or  by  dividing  the  number  of  square  inches 
of  belt  in  contact  with  the  pulley,  by  two  (2),  multiplying  this  quotient 
obtained  by  the  velocity  of  the  belt  in  feet  per  minute;  divide  this 
product  by  33000  and  the  quotient  will  be  the  HP  the  belt  will  transmit. 

Q.  89.  (1899-1900.)  Theoretically,  what  width  of  belt  would  be  re- 
quired to  transmit  500  HP,  with  driving  pulley  22  feet  in  diameter, 
turning  76  R.P.M.,  size  of  driven  pulley  9  feet,  the  belt  to  be  of  leather, 
two-ply?  From  a  practical  knowledge  of  belts,  what  width  would  you 
recommend? 

Ans.  89.    Velocity  in  feet  per  minute  equals — 

22  X  3.1416  X  76  =  5252.75  feet. 
5252.75  -5-  600  =  8.754  HP  per  1  inch  of  belt. 

•7-8,754  =  57  inches,  the  width  of  belt  required  under  the  con- 
aitions  of  the  question  for  the  transmission  of  500  HP. 
en  By  Different  rules  or  formulas  the  width  may  be  calculated  from 

«  H  it  w*     *8>  b-Ut  Under  good  Practical  conditions  and  circumstances 
a  belt  57  inches  in  width  will  be  ample  in  all  respects 

There  is  no  doubt  whatever  that  a  belt  50  inches  will  do  the  re- 
quired work  without  trouble  if  properly  cared  for 

PPOT,n™Vhi%rmmItt?®'S  opinion  that  a™ple  width  in  belting  is  more 
fn™ So  *\ working  too  close  to  the  width  that  will  actually  per- 
orm  the  work,  both  in  care  of  and  durability  of  the  belt. 

i  %  ^  <18"-1900.)  What  is  the  effective  pull  of  a  30  in  two-ply 
90  R.pr  Ml  '  transmittin*  30°  HP>  *i»  of  drive  pully  20  ft.  dia,"  turning 


Ans.  90.    The  velocity  in  feet  per  minute  equals  90  X  20  X  3.1416  — 
5654.88  feet. 

Then— 
300  X  33000 

=  1750.7  Ibs.,  the  effective  pull  upon  the  belt  under  condi- 


5654.86 
tions  stated  in  the  question. 

The  effective  pull  per  1  inch  of  width  of  the  belt  equal  1750.7  -=-  30 
=  58.356  pounds. 

Rule: — Reduce  the  horse-power  to  foot  pounds  and  divide  the  num- 
ber of  foot  pounds  by  the  velocity  of  the  belt  in  feet  per  minute  and 
the  result  will  be  the  effective  pull  upon  the  belt  in  pounds. 


Q.  95.  (1899-1900.)  At  what  point  in  the  stroke  of  an  engine  is  the 
pressure  the  greatest  on  the  guides? 

Ans.  95.  With  a  uniform  pressure  upon  the  piston,  the  maximum 
thrust  upon  the  guides  will  occur  when  the  cranks  are  at  right  angles 
to  the  axis  of  the  cylinder,  or  at  an  angle  of  90°. 

A  drop  in  pressure,  due  to  an  early  cut-off  or  other  causes,  would 
modify  this  statement  somewhat,  depending  upon  the  point  of  cut-off, 
and  the  angularity  of  the  rod.  The  maximum  thrust  upon  the  guides 
might  occur  at  a  point  earlier  in  the  stroke,  or  before  the  crank  had 
reached  its  greatest  angle. 

By  referring  to  sketch,  the  triangle  DEF  represents  the  crank  at  an 
angle  of  90°. 

The  indicator  diagram,  drawn  to  represent  an  initial  pressure  of 
100  pounds  absolute  scale,  60  pounds  per  inch  in  height,  with  a  25  per 
cent  of  the  length  of  the  stroke,  cut-off,  shows  that  when  the  piston 
has  reached  the  point  E,  representing  50  per  cent  or  one-half  stroke, 
and  the  crank  at  90°,  the  pressure  has  fallen  to  about  48  pounds  abso- 
lute. Consequently  the  maximum  pressure  will  take  place  at  a  point  in 
the  stroke  before  the  crank  has  reached  the  angle  of  90°  to  the  axle 
of  the  cylinder. 

Expressed  in  proportion  DE  :  DF  :  :  horizontal  thrust  :  the  down- 
ward thrust  on  guides.  Horizontal  thrust  equals  area  of  piston  multi- 
plied by  steam  pressure. 

The  distance  DE  is  found  by  the  rules  of  finding  the  sides  of  a  right 
angle  triangle,  hence — 

DE  =  the  sq.  root  of  the  difference  of  the  squares  of  the  height  on 
the  hypothenuse. 

Assuming  DF  to  equal  I,  the  ratio  of  the  length  of  the  crank  to  the 
connecting  rod  =  EF.  ' 

DE  =  1/EF2  —  DF8 

when  the  maximum  thrust  on  the  guide  at  E  equals — 
DF.  X  horizontal  thrust 


DE. 

(See  sketch.) 


Q.  96.  (1899-1900.)  What  will  be  the  greatest  thrust  on  the  gui'l-s 
(due  to  the  steam  pressure)  of  an  engine  24"x60",  having  a  connecting- 
rod  5y2  times  the  length  of  the  crank,  with  an  unbalanced  pressure  of 
50  Ibs.  per  square  inch  (gage)  on  the  acting,  or  working,  side  of  the 
piston? 

Ans.  96.     Engine  24"  X  60". 

Area  of  piston  =  242  X  .7854  =  452.39  sq.  in. 

Horizontal  thrust  parallel  with  the  axis  of  the  cylinder  =  452.30 
X  50  =  22619.50  pounds. 

Length  of  crank  =  60  -~  2  =  30  inches,  or  2.5  feet. 

99 


Lengrth  of  connecting  rod  equals  2.5  X  5.5  =  13.75  feet. 

Referring  to  sketch,  question  95. 

The  altitude  DF  of  the  triangle  DEF  equals  2.5  feet;   the  hypothe- 
nuse  EF  equals  13.75  feet.    The  base  DE  equals 

1/EF8— DF2  =1/13. 75a— 6.25s =1/182.8125= 13.52  +  feet, 
or  length  of  the  base  DE;  therefore 


2.5  X  22619.50 


13.52 
or  downward  thrust  on  the  guides. 


•  =  4182.6  pounds 


Q.  97.  (1899-1900.)  With  a  guide  5y2"  wide  and  shoe  of  cross-head 
18"  long,  what  is  the  greatest  thrust  on  the  shoe,  per  square  inch? 
Taking  the  same  conditions  as  stated  in  question  96. 

Ans.  97.  The  superficial  area  of  shoe  equals  18  inches  X  5^  inches 
=  99  sq.  in. 

As  found  in  Question  96,  the  total  thrust  upon  the  guides  is  4182.6 
pounds.  Then  the  total  thrust  per  sq.  inch  equals  4182.6  -4-  99  =  42.2 
pounds. 

Q.  98.  (1899-1900.)  In  comparison,  how  do  the  lengths  of  eonnect- 
ings  rods  affect  the  thrust  or  pressure  upon  the  guides? 

Ans.  98.  In  comparison  the  thrust,  or  pressure  exerted  upon  the 
guides  is  affected  inversely,  as  the  length  of  the  rods;  that  is,  the  longer 
the  connecting  rod,  as  compared  with  the  length  of  the  crank,  the  less 
will  be  the  pj  )ssure  upon  the  guide. 


Q.  99.  (1899-1900.)  If  the  diameter  of  a  piston  is  18  inches,  the 
pressure  at  the  beginning  of  the  stroke  is  130  Ibs.  (gage)  .per  square 
inch,  what  is  the  pressure  on  the  crank-shaft  at  this  point?  When  the 
crank  has  turned  to  an  angle  of  60°,  with  the  axis  of  the  cylinder,  the 
steam  pressure  has  dropped  to  115  Ibs.  gage  per  square  inch,  what  is 


the  pressure  on  the  crank-pin? 
described  by  the  crank-pin? 


What  is  the  force  tangent  to  the  circle, 


Ans.  99.  The  area  of  18  inch  piston  equals  182  X  .7854  =  254.47 
square  inches. 

Horizontal  thrust  at  beginning  of  stroke  equals  254.47  X  130  = 
33081.10  pounds. 

The  horizontal  pressure  on  crank-pin  when  at  an  angle  of  60°  to  the 
axis  of  cylinder  with  a  pressure  of  115  Ibs.  per  sq.  inch  equals  254.47 
X  115  =  29264.05  pounds. 

When  the  crank  is  in  a  position,  termed  dead  center,  that  is.  when 
piston  connecting  rods  and  crank  are  all  in  the  same  line,  the  horizontal 
thrust,  simply  produces  a  pressure  on  the  bearings  of  the  crank-shaft. 
When  the  crank  has  turned  to  an  angle  of  90°  or  at  right  angle  with 
the  axis  of  the  cylinder  the  pressure  of  the  steam  on  the  piston  is  all 
exerted  in  turning  the  shaft,  and  none  on  the  bearings;  this  is  because 
the  pressure  exerted  is  at  right  angles  to  the  crank,  or  tangent  to  the 
crank  circle. 

By  referring  to-the  diagrams,  when  the  crank  is  at  60°  with  the  axis 
of  the  cylinder,  the  pressure  is  exerted  in  the  direction  of  a  c  and  a  e. 
The  diagram  line  a  d  b  represents  the  horizontal  thrust  of  29264.05 
pounds. 

The  tangential  thrust  in  the  direction  a  c  will  be  as  a  b  :  a  f  •  • 
29264.05  :  X. 
Substituting. 

2"  :  1.73212  :  :  29264.05  :  X. 

X  =  25343.545  pounds,  equals  the  force  tangential  to  crank-pin 
circle  a  c. 

SEE   QUESTION   99. 

Note.—  The  sine  of  the  angle  of  60°  =  .86606.  Multiplying  the 
length  of  the  line  a  b  (2")  by  the  sine  of  the  60°  (.86606)  equals  the 
length  of  the  line  a  f  equals  1.73212  inches. 

The  force,  or  thrust,  in  direction  of  a  e  is  in  proportion   as  — 

ab  :  ad  :  :  29264.05  :  X 
Substituting— 

2"  :  1  :  :  29264.05  :  X 

X=  14632.025  pounds,  the  pressure  on  the  crank  shaft  when  the 
crank  is  at  an  angle  of  60°  with  the  axis  of  the  cylinder,  or  in  the 
direction  of  the  line  a  e. 

Note.—  The  cosine  of  the  angle  of  60°  equals  .5  Multiplying  the 
.length  of  the  line  a  b  (2")  by  the  cosine  of  the  angle  of  60°  (.5)  equals 
the  length  of  the  line  a  d  equals  1  inch. 

The  horizontal  thrust  is  represented  by  line  a  b. 
The  thrust  on  crank  shaft  at  60°  by  line  a  e 
The  thrust  tangential  at  60°  by  line  a  c. 

Q.  106.  (1899-1900.)  How  does  the  Corliss  type  of  engines  differ 
from  the  slide-valve  type  of  engines? 

excels™6  the  m°St   important  advantage   in   which   the   Corliss   type 
release>  by  the  mechanical  mechanism,   how  are  the  valves 


Ans   106.    The  Corliss  type  of  engine  differs  from  the  common  slide 
valve  type  in  that  it  has  four  valves,  one  at  each  end  of  the  cylinder 
Qe  admission  of  steam  and  one  at  each  end  of  the  cylinder  for  the 
exhaust  or  escape  of  the  steam. 

'he  steam  entering  the  cylinder  at,  or  near,  the  boiler  pressure  for 
?S!n?  P-  ,  he  Str°ke>  at  which  P°int  the  mechanism  that  opens 


>n      -  ,  > 

off  thP^T  V^lve^  is.released  and  allows  the  valves  to  close,  cutting 

remainder  of  the  work  to  be  performed 


type  excels  the  slide  valve  type  in  that  the  steam  is 
venienetX1ian:      ely>    c°nse(luently,    more    economically,    also    more    con- 
t  in  the  adjustment  of  the  valve  gear,  as  each  valve   is  inde- 


pendent  of  the  others,  less  friction  of  the  valve  gear  per  unit  of  power 
developed,  and  a  close  regulation  of  speed. 

After  the  valve  mechanism  is  released,  the  steam  valves  are  closed 
by  dash-pots,  or,  as  sometimes  called,  vacuum  pots,  consisting  of  a  cast 
iron  cylinder  fitted  with  an  air-tight  piston  attached  to  a  bell-crank 
on  the  valve  stem.  As  the  opening  mechanism  opens  the  valve  it  raises 
the  piston  in  the  dash  pot,  forming  a  partial  vacuum  on  the  under 
side  of  the  piston;  at  the  point  of  cut-off  the  opening  mechanism  is 
released  the  atmospheric  pressure  on  the  upper  side  of  this  piston 
closes  the  valve  rapidly.  To  prevent  pounding  a  little  air  is  admitted 
for  cushioning  the  piston  before  striking  the  bottom. 


Q.  107.  (1899-1900.)  An  engine  working  at  its  maximum  capacity, 
exhausting  to  the  atmosphere,  it  is  desired  to  increase  the  load  25  per 
cent,  without  an  increase  in  steam  pressure  or  the  speed,  how  could 
this  change  be  effected? 

Ans.  107.  When  an  engine  exhausts  to  the  atmosphere  the  steam 
which  fills  the  cylinder  at  the  end  of  the  stroke  has  to  be  forced  out 
against  the  atmospheric  pressure,  usually  reckoned  15  Ibs.  per  sq.  in. 

The  nature  of  steam  being  such  that  a  part  of  the  atmospheric 
pressure  can  be  removed. 

One  pound  of  steam  at  atmospheric  pressure  occupies  about  1642 
times  as  much  space  as  it  does  in  the  form  of  water. 

.  If  this  steam,  when  it  is  exhausted  from  the  cylinder,  is  allowed  to 
come  in  contact  with  cold  water,  or  a  cold  surface  within  a  vessel 
that  is  air-tight,  it  will  be  immediately  condensed,  occupying  about 
1/1600  of  its  original  volume,  creating  a  partial  vacuum  in  the  con- 
denser, the  water  and  air  being  removed  by  the  air  pump,  allowing 
the  piston  to  make  its  return  stroke  relieved  from  a  portion  of  the 
atmospheric  pressure  of  15  Ibs.  per  sq.  in. 

This  reduction  of  back  pressure  is  equal  to  a  corresponding  in- 
crease of  the  M.  E.  P.,  increasing  the  mean  efficiency  of  the  engine 
from  25  to  30  per  cent.  Therefore,  the  adding  of  a  condenser  will 
increase  the  capacity  of  the  engine  the  required  amount. 


Q.  108.  (1899-1900.)  In  good  practice,  how  marv  expansions  are 
advisable  in  a  single-cylinder  engine? 

Why  are  more  expansions  objectionable?  When  more  expansions 
are  desired,  how  can  they  be  obtained? 

When  the  engine  is  so  constructed  as  to  properly  give  more  expan- 
sion, what  is  this  type  of  engine  called? 

Ans.  108.  Usually  in  good  practice  it  is  desirable  to  have  from  3  to 
5  expansions  in  a  single  cylinder  engine;  an  allowance  of  from  20  to 
33  per  cent  cut-off,  which  is  about  the  economical  range  of  cut-off 
engines'. 

When  more  expansions  are  desired  it  is  more  practical  to  allow  the 
expansions  to  take  place  in  more  than  one  cylinder.  This  type  of 
engine  is  called  a  compound  engine. 


Q.  109.  (1899-1900.)  When  the  expansion  takes  place  in  more  than 
one  cylinder,  for  example,  a  two-stage  compound,  how  are  the  number 
of  expansions  found,  and  are  they  effected  by  the  cut-off  in  the  low 
pressure  cylinder? 

Ans.  109.  When  the  expansions  take  place  in  a  two-stage  compound 
the  number  of  expansions  are  found  by  dividing  the  initial  absolute 
pressure  in  the  H.  P.  cylinder  by  the  absolute  terminal  pressure  in  the 
L.  P.  cylinder,  or  it  is  the  product  of  &,  number  of  expansions  in  the 
H.  P.  cylinder  by  the  number  of  expansions  in  the  L.  P.  cylinder. 

103 


Formula— 

—  the  number  of  expansions  of  the  two  cylinders. 

t    t1 

In  which— 

T  =  absolute  initial  pressure  H.  P.  cyl. 

t  —  absolute  terminal  pressure  H.  P.  cyl. 

t*=i  absolute  terminal  pressure  L.  P.  cyl. 

The  number  of  expansions  are  not  affected  by  the  point  of  cut-off 
in  the  low  pressure  cylinder. 


Q.  110.  (1899-1900.)  A  compound  engine,  working  with  an  initial 
pressure  of  130  Ibs.  absolute  pressure  expanded  to  a  terminal  pressure 
of  26  Ibs.  absolute  in  the  H.  P.  cylinder;  then  received  and  expanded 
to  a  terminal  pressure  of  8  Ibs.  absolute  in  the  L.  P.  cylinder;  find  the 
total  number  of  expansions  in  the  two  cylinders. 

Ans.  110.     130  pounds  absolute  =  initial  pressure  H.  P.  cyl. 

26  pounds  absolute  =  terminal  pressure  H.  P.  cyl. 
8  pounds  absolute  =  terminal  pressure  L.  P.  cyl. 

The  number  of  expansions  taking  place  in  high  press,  cyl.  =  130  -f- 
26  =  5  expansions. 

Low  pressure   cyl.  =  26  -=-  8  =  3.25   expansions. 

Whole  number  of  exp.  =  5  X  3.25  =  16.25  expansions. 

By  formula — 

T    t 

=  total  number  of  expansions. 

t   t1 

130       26       130 

Substituting X  —  = =  161/4  total  exp. 

26         8          8 


Q.  111.  (1899-1900.)  For  what  purpose  is  a  receiver  placed  between 
the  two  cylinders  of  a  compound  engine?  What  should  be  its  relative 
capacity,  and  to  which  cylinder  of  the  engine  does  the  relation  exist? 

Does  the  disposition  or  arrangement  of  the  cranks  affect  the  size  of 
the  receiver? 

What  conditions  would  be  favorable  for  the  non-use  of  a  receiver, 
the  steam  passing  through  the  exhaust  pipe  from  the  high  to  low 
pressure  cylinder? 

Ans.  111.  A  receiver  consists  of  an  enclosed  vessel  or  cylinder,  gen- 
erally provided  with  a  steam  jacket  or  some  means  for  reheating  the 
steam  and  to  provide  against  loss  of  heat  from  radiation,  etc.  They  are 
sometimes  called  re-heaters. 

The  H.  P.  cylinder  exhausting  into  this  receiver  and  the  L.  P. 
cylinder  taking  its  supply  from  it. 

It  provides  for  a  volume  of  steam  close  at  hand  to  supply  the  low 
pressure  cylinder  without  a  serious  reduction  or  drop  in  the  pressure, 
and  at  a  low  maximum  velocity. 

The  volume  of  the  receiver  is  usually  from  1  to  1.75  times  the 
volume  of  the  H.  P.  cylinder. 

The  larger  ratio  is  more  favorable  to  a  straight  back  pressure  line 
of  the  H.  P.  cylinder  and  steam  line  of  the  L.  P.  cylinder  indicator 
cards. 

The  angle  that  the  cranks  are  set  to  each  other  does  affect  the 
required  volume  of  the  receiver. 

The  most  favorable  condition  for  the  non-use  of  a  receiver  is  when 
the  cylinders  are  placed  in  line  with  each  other  and  both  pistons  are 
attached  to  the  same  crank,  as  a  tandem  compound,  or,  when  the  cranks 
are  set  at  an  angle  of  180°,  or  nearly  so,  to  each  other. 

104 


Q.  112.  (1899-1900.)  Give  the  rule  to  find  the  receiver  pressure 
necessary  to  balance  the  load  on  an  engine? 

What  receiver  pressure  is  required  to  balance  the  load  on  a  28"  X 
52"  X  72"  cross-compound  engine,  the  steam  pressure  140  Ibs.  absolute; 
vacuum  maintained  at  28  inches? 

Ans.  112.  In  order  that  the  work  of  a  compound  engine  may  be 
divided  equally,  or  nearly  so,  the  receiver  pressure  should  be  propor- 
tional to  the  initial  pressure  of  the  H.  P.  cylinder,  according  to  the 
ratio  of  the  area  of  the  piston  of  the  H.  P.  cylinder  and  the  area  of  the 
piston  of  the  L.  P.  cylinder. 

Rule.— To  find  the  proper  pressure  to  maintain  in  the  receiver  of  a 
compound  to  balance  the  load. 

Form  the  proportion: 

As  the  area  of  the  H.  P.  piston  is  to  the  area  of  the  L.  P.  piston 
so  is  the  required  receiver  pressure  to  the  initial  pressure  less  the 
receiver  pressure  (which  acts  as  a  back  pressure). 

As  the  areas  of  circles  vary  directly  as  the  square  of  their  diameters, 
it  is  not  necessary  to  find  the  areas  of  the  pistons  use  the  square  of  the 
diameters  instead. 

Example — 

Compound  28"  X  52"  X  72". 

Initial  press.  140  Ibs.  ab. 

X  =  the  receiver  pressure. 

Then— 
.     282  :  522  :  :  x  :  (140  — x) 

=  784  :  2704  :  :  x  :  (140  — x) 

The  product  of  the  extremes  =  784  (140  —  x) 

The  product  of  the  means  =  2704  x  or  2704  x  =  784  (140  — x)  784 
(140  —  x)  =  109760  —  784  x. 

Then  2704  x  =  109760  — 784  x  =  3488  x  =  109760. 
109760 

X  = =31.47  Ibs.  ab.  receiver  pressure. 

3488 

28"  vacuum  =  28  X  .491  =  13.7  pounds. 

31.47  — 13.7  =  17.77  pounds  receiver  pressure,  as  shown  by  the  gage. 


Q.  113.  (1899-1900.)  In  good  practice,  how  should  the  work  be  dis- 
tributed on  a  cross-compound  engine?  How  is  this  distribution  brought 
about? 

Ans.  113.  The  work  should  be  so  distributed  that  each  cylinder  will 
perform  an  equal  share  of  the  same;  that  the  initial  strains  in  each 
cylinder  sb^"M  be  as  near  the  same  as  possible.  That  the  drop  in 
pressure  between  the  high  pressure  cylinder  and  the  receiver  should  be 
as  small  as  possible. 

This  distribution  is  brought  about  by  the  adjustment  of  the  cut-off 
in  the  L.  P.  cylinder,  shortening  up  the  cut-off  increases  the  receiver 
pressure,  allowing  the  low  pressure  cylinder  to  perform  a  greater  pro- 
portion of  the  work. 

Lengthening  out  the  cut-off  has  a  vice  versa  effect. 


Q.  114.  (1899-1900.)  What  are  the  advantages  derived  from  the 
use  of  compound  engines  over  the  use  of  simple  engines? 

What  conditions  in  practice  are  necessary  for  the  best  economy  in 
the  use  of  multiple  cylinder  engines,  condensing  or  non-condensing? 

Ans.  114.  The  advantages  of  dividing  the  expansion  among  two  or 
more  cylinders,  or  the  use  of  multi-cylinder  engines  are  the  use  of 
higher  initial  pressures,  a  wider  range  of  expansion,  with  a  minimum 

105 


of  cylinder  condensation,  avoiding  excessive  strains  on  the  metal  by 
sudden  expansion  and  contraction. 

Sizes  of  cylinders  adapted  to  the  work  to  be  performed:     A  con- 
stant steam  pressure  and  a  constant  load  are  required  for  economy. 


Q.  115.  (1899-1900.)  Upon  what  depends  the  ratio  for  capacity  of 
the  several  cylinders  of  multiple  cylinder  engines?  Explain  how  the 
ratio  is  found. 

Ans.  115.  The  ratio  of  capacity  of  the  cylinders  depends  upon  the 
range  of  temperature  or  pressure  desired,  or  the  ratio  of  the  initial 
pressure  to  that  of  the  desired  terminal  pressure  in  the  following 
cylinder;  this  is  the  relation  that  the  volume  of  the  H.  P.  cylinder  bears 
to  the  L.  P.  cylinder. 

Rule. — Divide  the  absolute  initial  pressure  of  the  first  cylinder  by 
the  absolute  terminal  pressure  of  the  last  cylinder  and  multiply  this 
quotient  by  the  desired  per  cent  of  cut-off  in  first  cylinder. 


Q.  116.  (1899-1900.)  What  should  be  the  ratio  of  capacity  of  the 
cylinders  of  an  engine  using  steam  at  an  initial  pressure  of  135  Ibs. 
absolute  per  square  inch,  expanding  to  a  terminal  pressure  of  10  Ibs. 
absolute  in  the  L.  P.  cylinder,  the  point  of  cut-off  30  per  cent  of  the 
length  of  the  stroke? 

Ans.  116.     135  Ibs.  absolute  =  initial  pressure. 

10  Ibs.  absolute  =  terminal  desired  in  L.  P.  cyl. 

30%  —  cut-off. 

Then — 

135  -r- 10  =  13.5   expansions. 

13.5  X  30  =  4.05  ratio  of  capacity. 

Or,  -.1  :  4.05  =  ratio  of  capacity  of  the  H.  P.  cyl.  :  L.  P.  cyl. 


Q.  117.  (1899-1900.)  Give  rule,  or  formula,  for  finding  the  approx- 
imate H.  P.  of  a  compound  engine. 

Ans.  117.  Rule  for  finding  the  approximate  H.  P.  of  a  compound 
engine — 

Assume  that  the  entire  work  is  to  be  done  and  the  expansion  all 
taking  place  in  the  L.  P.  cylinder. 

Then,  1  +  hyp.  log.  of  the  total  number  of  the  expansions,  multi- 
plied by  the  terminal  pressure  in  the  L.  P.  cylinder  will  equal  the 
average  mean  effective  pressure,  due  to  the  expansions;  this  product 
multiplied  by  the  area  of  the  L.  P.  cylinder's  piston  and  by  the  piston 
speed  in  feet  per  minute;  this  product  divided  by  33,000  will  give  the 
approximate  H.  P.  of  the  engine. 

Formula — 

H.  P.  =  horse-power  of  compound  engine. 

H  =  *  +  hyp.  log.  of  total  number  of  expansions. 
—  terminal  pressure,  L.  P.  cylinder. 

®  =  average  M.  B.  P.  due  to  number  of  expansions. 

b  =  piston  speed  in  ft.  per  minute. 

A  =  area  of  piston. 

E  =  T.  H. 

Then— 

A.S.E. 

H.  P.  = 

33,000 

106 


Q.  118.  (1899-1900.)  Required  size  of  cylinders  for  a  compound 
engine,  condensing  of  2,000  H.  P.,  5  feet  stroke  and  72  R.  P.  M.,  or  720 
piston  feet  per  minute. 

Boiler  pressure  140  Ibs.  absolute,  an  allowance  of  5  Ibs.  for  drop  be- 
tween the  boiler  and  throttle. 

Terminal  pressure  L.  P.  cylinder  6  Ibs.  per  square  inch.  Back  pres- 
sure 3  Ibs.  per  square  inch. 

Prove  the  work  by  rule  found  in  Question  117. 

Ans.  118.  The  theoretical  diameter  for  a  two-stage  compound  of 
2,000  H.  P.,  60-inch  stroke,  72  R.  P.  M. 

Boiler  pressure  140  Ibs.  absolute. 

Terminal  pressure  6  Ibs. 

Back  pressure  3  Ibs. 

Drop  in  pressure  between  boiler  and  throttle  5  Ibs. 

140  —  5  =  135  Ibs.  initial  pressure  absolute  at  throttle. 

135  -j-  6  =  22.5  total  number  of  expansions. 

The  expansion  in  each  cylinder  is  equal  to  V22.5=4.74. 

The  terminal  pressure  in  the  H.  P.  cyl,  is  equal  to  135  H-  4.74  = 
28.48  Ibs.  absolute. 

Average — 

M.  E.  P.  of  H.  P.  cylinder  is  equal  to  (1  +  hyp.  log.  of  4.74)  28.48 
=  72.79  Ibs.  absolute. 

72.79  —  28.48  =  44.31  Ibs.  =  M.  E.  P.  of  H.  P.  cyl. 

Average — 

M.  E.  P.  of  L.  P.  cylinder  is  equal  to  (1  +  hyp.  log.  of  4.74)   6  =  15.34 
Ibs.   absolute. 

15.34  —  3  =12.34  pounds.    M.  E.  P.  of  L.  P.  cylinder. 

It  is  desirable  to  have  the  cylinders  so  proportioned  that  the  work, 
with  the  above  conditions,  one-half  of  which,  to  wit,  1,000  H.  P.,  be 
performed  in  each  cylinder. 

Then,  the  area  of  the  low  pressure  cylinder  will  equal 
33,000  X  H.  P. 


piston  speed  in  ft.  X  average  M.  E.  P. 
Substituting — 

33,000  X  1,000         33,000,000 

= =3,714.21  sq.  in. 

720  X  12.34  8,884.8 

the  area  of  L.  P.  cyl. 

V3.714.21 


=  68.70  in.,  diam.  of  L.  P.  cyl. 


.7854 

Area  of  the  high  pressure  cyl.  will  equal — 
33,000  X  1,000         33,000,000 

= =1,034,379    sq.   in. 

720  X  44.31  31,903.2 


V1'034'3;9=36.29in..dia.ofH.P.cyl. 

.7854 

The  size  of  the  compound  engine  as  per  the  conditions  stated  in 
question  will  be— 36"  X  69"  X  60"— at  72  R.  P.  M.  with  140  Ibs.  absolute 
boiler  pressure,  22.5  expansions,  or  4.74  expansions  in  each  cylinder 
exhausting  into  condenser  against  3  Ibs.  back  pressure,  or  into  a 
vacuum  of  14.7  —  3  =  11.7  Ibs. 

11.7  Ibs.  X  2.036  =  23.82  in.  vacuum. 

692 
The   ratio  of  cylinder  =  —  =  3.67. 

362 

The  engine  under  the  above  conditions  will  develop  2,000  H.  P. 
Proof  by  rule  foun.d  in  Ans.  117. 

107 


Whole  number  of  expansions  =  135  -  6  =  22.5,    (1  +  hyp.  log.    of 
22.5)  =4.11353. 

6  X  4.11353  =  24.68118  Ibs.  effective  pressure. 
Area  L.  P.  piston  =  3,729.38  sq.  in. 
Speed  of  piston  =  720  ft.  per  min. 
Average  effective  pressure  =  24.68  Ibs. 
Tnen — 

3,729.38  X  720  X  24.68 

J -  =  2,012.05    H.    P., 

33,000 
a  difference  only  of  12.05  horse-power. 


Q  119.  (1899-1900.)  Describe  the  manner  of  attaching  the  indi- 
cator to  an  engine.  What  care  should  be  taken  in  such  preparation 
for  the  attachment?  How  to  obtain  the  drum-motion  and  what  care 
should  be  taken? 

Describe  the  manner  of  taking  diagram,  and  note  the  important 
data  usually  taken. 

Ans.  119.  The  indicator  is  attached  to  the  end  of  the  cylinder, 
through  holes  drilled  and  tappec1  tor  y2"  pipe,  into  the  side  or  top, 
whichever  is  the  most  convenient,  to  connect  with  a  reducing  motion, 
attached  to  the  crosshead. 

The  holes  must  be  drilled  in  such  a  position  that  they  will  com- 
municate with  the  clearance  space  and  not  become  covered  by  the 
piston  when  it  is  at  the  extreme  end  of  the  cylinder;  the  indicator 
should  be  attached  as  direct  as  possible  to  the  cylinder. 

In.  most  engines  the  holes  are  drilled  and  tapped  in  the  shop  before 
it  is  sent  out.  Some  are  not,  however,  and  in  those  that  are  not  care 
should  be  taken  to  locate  the  holes  and  to  keep  the  chips  from  the 
drill  from  getting  into  the  cylinder,  especially  the  front  end. 

As  the  length  of  the  diagram  represents  the  travel  of  the  piston  for 
%  revolution,  or  one  stroke  of  the  engine,  and  the  length  of  this  dia- 
gram is  determined  by  the  rotation  of  the  paper  drum,  this  motion 
must  be  taken  from  some  part  of  the  engine  which  has  a  motion  coin- 
ciding with  that  of  the  piston;  the  crosshead  is  the  most  convenient 
part  for  this  purpose. 

This  drum  motion,  or  reducing  motion,  as  it  is  called,  is  obtained 
by  the  Brumbo  pulley,  or  the  pantograph,  reducing  wheels,  and  nu- 
merous other  devices,  so  long  as  they  reduce  the-  motion  proportion- 
ately to  the  stroke  of  the  engine  and  parallel  with  the  crosshead  or 
piston;  so  that  when  the  crosshead  has  made  *4  of  its  stroke  the  pencil 
will  have  traveled  %  the  length  of  the  diagram,  and  so  on. 

In  taking  diagrams  from  engines,  the  indicators  should  be  thor- 
oughly warmed  up  by  opening  the  steam  cock  between  the  indicator 
and  the  cylinder,  the  working  parts  should  work  very  smooth,  the 
spring  selected  to  give  the  diagram  should  be  one  that  will  make 
a  diagram  from  1%  to  2  inches  high;  dividing  steam  pressure  per 
gage  by  about  1%  gives  a  scale  of  spring  that  makes  a  very  good 
height  for  diagram. 

The  pencil  should  be  sharp,  smooth,  and  so  adjusted  that  when  the 

sssure  is  applied  it  will  make  a  smooth,  fine,  distinct  line.  The  cord 
leading  from  the  crosshead  or  reducing  motion  to  the  indicator  should 
be  so  adjusted  that  the  paper  drum  in  rotating  will  not  bring  up 
against  the  stops. 

After  placing  the  paper  around  the  drum  taut  and  smooth  start 
the  paper  drum  in  motion,  open  the  shutoff  cock,  allowing  steam  on 
the  under  side  of  the  piston.  Press  the  pencil  against  thf  pape?  for 
one  revolution  or  more,  then  shut  off  the  steam  to  the  indicator  and 

1 08 


again  press  the  pencil  to  the  paper  and  draw  the  atmospheric  line, 
which  is  drawn  with  the  atmospheric  pressure  on  both  sides  of  the 
piston. 

As  much  of  the  following  data  as  c°u  should  be  obtained  and  noted 
on  the  diagram:  The  date,  the  hour  of  raking  the  diagram,  the  type  or 
make  of  the  engine,  which  end  of  cylinder  and  which  engine  if  one 
of  a  pair,  the  diameter  of  cylinder,  and  length  of  stroke,  the  diameter 
of  the  piston  rod,  the  number  of  R.  P.  M.,  the  steam  pressure,  by 
gage,  at  the  boilers  and  at  the  engine,  atmospheric  pressure  by  barom- 
eter, the  vacuum,  by  gage,  in  the  condensers  (if  the  engine  is  a  con- 
densing type),  temperature  of  feed  water  at  boilers,  the  receiver 
pressure  by  gage  if  the  engine  is  a  compound,  if  not  note  the  back 
pressure  per  gage,  and  the  scale  of  spring  used  in  taking  the  diagram; 
also  note  any  other  matter  connected  with  the  plant  that  may  come 
to  your  notice. 


Q.  120.  (1899-1900.)  What  is  the  value  of  the  indicator  diagrams 
for  the  successful  operation  of  a  steam  plant? 

Describe  the  workings  and  use  of  a  planimeter. 

Ans.  120.  The  indicator  diagrams  are  the  result  of  two  motions — 
thus:  The  horizontal  of  the  paper  in  exact  correspondence  with  that 
of  the  piston,  it  represents  the  stroke  of  the  engine  on  a  reduced  scale, 
and  the  vertical  movement  of  the  pencil  in  exact  ratio  to  the  steam 
pressure,  exerted  in  the  cylinder;  consequently,  it  represents  by  its 
length  the  stroke  of  the  engine,  by  its  height  at  any  point,  the  pres- 
sure on  the  piston,  at  a  corresponding  point  in  the  stroke. 

The  shape  of  the  diagram  depends  very  much  upon  the  manner  in 
which  the  steam  is  admitted  to  and  released  from  the  cylinder  of  the 
engine.  It  shows  the  arrangement  of  the  valves  for  admission  of 
steam,  cutoff,  release  and  compression  of  the  steam. 

The  adequacy  of  the  ports  and  passages  for  admission  and  exhaust, 
and  when  applied  to  the  steam  chest,  show  the  adequacy  of  the  steam 
pipe. 

The  amount  of  power  developed,  the  quantity  lost  in  various  ways 
— by  wire  drawing,  back  pressure,  premature  release,  or  any  other 
maladjustment  of  the  valves. 

It  has  proved  itself  very  useful  to  the  designers  of  steam  engines 
in  figuring  out  the  distribution  of  the  horizontal  pressure  on  the  crank 
pin,  the  angular  distribution  of  the  tangential  components  of  the  hori- 
zontal pressure;  in  other  words,  the  rotative  effect  around  the  path  of 
the  crank. 

Taken  in  combination  with  a  measurement  of  feed  water  and  coal 
used,  the  economy  of  the  plant  can  be  found. 

The  degree  of  excellence  to  which  the  steam  engine  of  to-day  has 
been  brought  is  due  principally  to  the  use  of  the  indicator,  as  a  careful 
study  of  the  diagrams  and  different  conditions,  load,  pressure,  etc., 
is  the  only  means  of  becoming  familiar  with  the  action  of  steam  in 
the  cylinder  of  an  engine.  The  indicator  furnishes  many  other  items 
of  importance  when  the  economy  of  the  generation  and  the  use  of 
steam  is  to  be  considered. 

The  planimeter  is  an  instrument  designed  to  measure  the  areas  of 
irregular  figures.  It  is  operated  by  moving  a  tracer,  with  which  it  is 
fitted,  over  the  lines  of  the  diagram  or  figure  to  be  measured,  and 
records  the  area  upon  a  graduated  wheel  and  Vernier  scale.  Some 
of  the  instruments  are  made  to  read  the  M.  E.  P.  or  the  I.  H.  P.  direct 
from  the  wheel  without  the  use  of  any  figures.  In  measuring  the  I. 
H.  P.  of  cards  much  time  and  labor  can  be  saved  by  the  use  of  the 
planimeter. 

109 


Q  1  (1900-1901.)  Give  three  ways  of  establishing  the  theoretical 
expansion  curve  on  an  indicator  diagram,  accompanying  answer  with 
a  sketch  of  each  method. 

Ans.  1.  To  establish  the  theoretical  curve  it  is  well  to  remember  that 
the  pressure  of  steam  varies  inversely  as  its  volume;  from  this  con- 
sideration we  can  readily  locate  any  number  of  points  in  the  curve. 

Again,  all  pressures  must  be  measured  from  perfect  vacuum,  and 
clearances  must  also  be  taken  into  account. 

By  referring  to  the  sketches  we  find  the  vacuum  line  denoted  by 
V,  and  the  clearance  line  by  C;  the  vacuum  line  laid  off  by  the 
scale  of  the  spring,  14.7  pounds  below  the  atmospheric  line,  and  the 
clearance  line  C  the  distance  h  from  the  beginning  of  the  diagram, 
representing  the  added  length  of  the  cylinder  that  would  represent 
or  equal  the  capacity  of  the  clearance  space  on  one  end  of  the 
cylinder. 

There  are  several  ways  of  establishing  the  points  in  the  theoretical 
curve.  The  curve  may  be  drawn  from  any  point  in  the  real  curve, 
the  only  restriction  being  that  the  point  must  be  selected  when  the 
steam  valve  and  the  exhaust  valve  are  closed. 

The  following  are  several  methods  of  establishing  the  theoretical 


Sketch  ~/ 


Through  the  point  of  release  b,  draw  any  line  as  aB  and  make 
AB  equal  to  ab.  Then  draw  any  other  line  as  GD,  and  Gd,  equal  to 
AD,  then  will  d  be  a  point  in  the  curve  bA,  as  shown  by  the  dotted 
lines  above  the  real  curve.  By  drawing  a  number  of  lines  through 
the  point  A,  and  following  the  same  method  in  regard  to  laying  off 
the  distances  any  number  of  points  in  the  curve  may  be  located. 

The  following  geometrical  method  for  locating  the  points  in  the 
theoretical  curve  is  perhaps  the  most  used: 

Select  any  point  as  I  in  the  actual  curve  and  from  this  point  draw 
a  vertical  line  to  J,  on  the  line  B;  the  line  B  representing  the  boiler 
pressure,  or,  it  may  be  a  line  drawn  at  any  convenient  height. 

From  the  point  J  draw  a  line  to  the  point  K,  K  being  the  inter- 
section of  the  vacuum  and  clearance  line;  from  I  draw  the  line  IL, 
parallel  with  the  atmospheric  line  and  at  right  angles  to  IJ. 

From  L  the  point  of  intersection  on  the  diagonal  line  JK  and  the 
horizontal  line  IL  draw  the  vertical  line  LM. 

The  point  M  is  the  point  of  theoretical  cut-off.  Fix  upon  any 
number  of  points  as  1,  2,  3,  4,  5  on  the  line  B  and  draw  from  these 
points  lines  to  the  point  K;  from  the  intersections  of  these  lines  with 
the  line  LM  draw  horizontal  lines,  and  from  points  1,  2,  3,  etc.,  draw 

no 


vertical  lines  parallel  to  IJ.  The  intersection  of  these  horizontal  and 
vertical  lines  will  establish  the  desired  points  in  the  theoretical 
curve. 

On  the  diagram  draw  vertical  lines,  commencing  at  the  clearance 
line   spacing  equal   distances   apart   and   number   them   as   shown   in 


sketch.  It  is  not  necessary  to  pay  any  attention  to  coming  out  even 
with  the  end  of  the  diagram,  and  the  more  numerous  these  vertical 
lines  are  the  greater  the  accuracy  in  developing  the  curve. 

Selecting  the  point  14  on  the  line  V,  the  pressure  represented  by 
the  scale  of  the  spring,  we  find  to  be  18  pounds  absolute. 

The  number  of  volumes  is  according  to  the  units  adopted  and 
represented  by  the  distance  apart  of  the  vertical  line  14,  hence  to 
find  the  pressure  for  any  other  line: 

Multiply  the  pressure  by  the  number  of  lines  and   divide  by  the 


number  of  lines  upon  which  it  is  desired  to  determine  the  pressure. 
For  example,  18  X  14  =  252.  If  we  divide  this  by  7,  for  instance,  it 
equals  36  pounds  to  be  set  off  by  the  scale  of  the  diagram  on  line  7. 
By  following  out  the  above  method  and  connecting  the  points 
together  the  theoretical  curve  for  the  diagram  will  be  established. 


Q.  2.  (1900-1901.)  Of  what  advantage  is  the  theoretical  curve 
and  what  does  it  indicate  or  represent?  Also  describe  the  adiabatic 
and  isothermal  curves.  Which  one  is  used  in  plotting  the  theoretical 
curve?  Why? 

Ans.  2.  The  defects  in  engines  as  revealed  by  the  indicator  diagram 
are  most  clearly  understood  by  comparing  the  diagram  of  the  engine  in 
question  with  a  theoretically  perfect  diagram — that  is,  one  which 
would  be  obtained  from  an  engine  in  which  both  the  designs  and  the 
adjustments  were  perfect  in  every  way. 


Adiabatic  curve  is  defined  as  the  curve  of  no  transmission  of  heat. 
Isothermal  curve  is  defined  as  the  curve  of  equal  temperature  and 
is  a  common  or  rectangular  hyperbola. 

The  isothermal  curve  is  usually  used   in   plotting  the  theoretical 

The  reason  why  this  curve  is  usually  used  is  that  according  to 
"Mariotte's"  law  the  volume  of  a  perfect  gas,  the  temperature  being 
kept  constant,  varies  inversely  as  its  pressure,  and  in  calculations 
of  the  mean  pressure  of  expanded  steam  in  engines  it  is  generally 
assumed  that  steam  expands  according  to  "Mariotte's"  law,  the  curve 
of  the  expansion  line  being  a  hyperbola. 


Q.  3.  (1900-1901.)  Explain  fully  how  you  ascertain  by  card  the 
number  of  pounds  of  water  consumed  per  indicated  horse  power  (I. 
H.  P.)  of  an  engine,  either  single  high  pressure  or  compound  con- 
densing. 

Ans.  3.  The  indicator  diagram  enables  us  to  find  approximately 
the  amount  of  steam  consumed  by  the  engine. 

The  rule  for  calculating  the  number  of  pounds  of  steam  consumed 
by  an  engine  per  horse-power  is  as  follows: 
Rule- 
Take  two  points,  one  on  the  expansion  line  before  release  and  one 
on  compression  line  equally  distant  from  the  vacuum  line.     Find  by 
the  scale  of  the  spring  used  the  pressure  of  steam  at  these  points,  and 
from  the  steam  table  find  the  weight  of  a  cubic  foot  of  steam  at  that 
pressure.    Multiply  this  weight  by  the  distance  between  the  two  points 
and  by  the  constant  13750. 

Divide  the  product  by  the  M.  E.  P.  and  by  the  length  of  the  dia- 
gram. 

The  result  will  be  the  number  of  pounds  of  steam  consumed 
per  I.  H.  P.  per  hour,  as  shown  by  the  diagram. 

This  computed  consumption  of  steam  is  for  various  reasons  always 
less  than  the  actual  consumption. 

The  difference  between  the  theoretical  water-consumption  found 
by  the  rule  and  the  actual  consumption  as  found  by  test  represents 
"water  not  accounted  for  by  the  indicator"  due  to  cylinder  condensa- 
tion, leakage  through  ports,  radiation,  etc. 

In  relation  to  multi-cylinder  engines  the  supposition  is  that  the 
work  is  done  in  one  cylinder. 


Q.  4.  (1900-1901.)  Explain  the  various  methods  of  obtaining  the 
mean  effective  pressure  (M.  E.  P.)  from  indicator  cards. 

Ans.  4.  In  general  practice,  the  M.  E.  P.  is  found  by  the  use  of  a 
planimeter  or  by  computing  by  the  use.  of  ordinates.  It  can  also  be 
found  by  plotting  on  cross-section  paper. 

The  planimeter,  as  has  been  heretofore  stated,  is  an  instrument 
by  which  the  area  of  irregular  surfaces  may  be  accurately  measured. 
Some  instruments  are  so  constructed  as  to  read  the  M.  E.  P.  direct 
while  others  give  the  area  only.  Having  the  area  given  the  M  E  P' 
is  found  by  multiplying  the  area  in  square  inches  by  the  scale  of 
the  spring  and  dividing  this  product  by  the  length  of  the  diagram. 

ihe  computing  by  the  means  of  ordinates  is  quite  generally  under- 
stood without  much  explanation;  the  total  number  of  ordinates— that 
is,  their  length— divided  by  the  total  number  equals  their  mean  length 
m  inches.  This  multiplied  by  the  scale  of  the  spring  equals  the 


Q.  32.  (1900-01.)  How  would  you  determine  if  the  steam  pipe  lead- 
ing to  your  engine  was  large  enough,  that  is,  of  sufficient  diameter? 

Ans.  32.  The  sufficient  diameter  of  steam  pipe  for  the  work  of  the 
engine  could  be  determined  by  the  use  of  the  indicator  both  on  the 
engine  cylinders  and  on  the  pipe  proper. 

If  in  indicating  an  engine,  the  drop  of  initial  pressure,  the  falling 
off  of  the  steam  line,  would  prove  conclusively  that  there  was  too  con- 
tracted area  in  the  steam  passages  of  the  engine  or  that  the  pipe  was 
too  small,  or  both,  to  locate  the  trouble  the  indicator  can  be  placed 
on  the  steam  chest;  also  on  the  pipe  and  a  diagram  taken  while  the 
engine  is  working  at  its  usual  or  rated  load.  The  sudden  falling  off  of 
the  pressure  would  indicate  the  trouble. 


Q.  34.  (1900-01.)  Give  in  a  general  way  your  practice  for  using  and 
handling  of  valves.  Also  the  best  practice  as  to  kinds  of  valves  to  use 
and  not  to  use. 

Ans.  34.  A  general  rule  for  locating  and  handling  valves  is  to  avoid 
forming  water  pockets.  Junction  valves  should  never  be  placed  close 
to  the  boiler  if  the  main  steam  pipe  is  above  the  boiler,  but  put  it  on 
at  the  highest  point  of  the  junction  pipe. 

Never  let  a  junction  pipe  run  into  the  main  at  the  bottom,  but  into 
the  top  or  side. 

Always  use  an  angle  valve  when  convenient,  as  there  is  more  room 
in  it. 

Ail  gate  valves  of  size  4  inches  and  upwards  should  be  by-passed; 
this  saves  the  gate  and  seat  of  valve  from  wear  and  is  also  convenient 
in  equalizing  the  pressure  before  opening  the  large  valve.  In  cutting 
in  a  boiler  a -by-pass  will  oe  found  very  convenient  in  equalizing  the 
pressure  before  opening  the  main  junction  valve.  Blow-off  valves 
should  be  opened  wide  open  to  allow  sediment  to  pass  out  and  not 
lodge  in  the  valve,  thus  ruining  it  when  closed.  Globe  valves  should 
not  be  used  on  indicator  pipes. 

For  water  always  use  gate  valves,  or  angle  valves,  or  stop-cocks. 

Valves  with  renewable  discs  are  recommended  in  some  cases.  It  is 
also  recommended  that  in  case  of  inspection  of  a  boiler  connected  to 
others  working,  as  a  precaution,  that  the  junction  pipe  be  disconnected 
and  a  blank  flange,  or  a  blank  be  inserted  between  the  flanges,  making 
the  boiler  surely  safe  for  inspection  or  cleaning. 


Q.  36.  (1900-01.)  For  what  purpose  is  a  separator  placed  on  a 
steam  pipe,-  and  for  the  greatest  efficiency  where  should  it  be  placed? 

Ans.  36.  The  function  of  a  separator  is  to  remove  from  vapors  and 
gases  the  liquids  and  solids  which  they  carry  along  with  them. 

Thus  a  live  steam  separator  is  intended  mainly  to  remove  the  en- 
trained water  from  steam,  while  the  separator  placed  in  the  exhaust 
piping  when  the  exhaust  steam  is  to  be  condensed  and  again  used  in 
the  boilers  is  intended  to  remove  the  grease  which  it  has  accumulated 
in  its  passage  through  the  engine. 

The  use  of  dry  steam  in  an  engine  cylinder  is  very  important,  not 
only  because  an  accumulation  of  water  is  a  menace  to  safety,  but  also 
because  entrained  water  involves  a  very  considerable  reduction  in  the 
economy  of  operation. 

The  water  carried  into  the  steam  cylinder  not  only  carries  away 
heat  from  the  boiler  which  is  incapable  of  doing  any  work  in  the  engine, 
but  it  also  materially  increases  the  initial  condensation,  one  of  the 
most  important  losses  of  power  in  the  engine. 

Hence  there  should  be  some  device  provided  to  ensure  dry  steam, 


The  principles  upon  which  the  action  of  steam  separators  should  be 
based  are  as  follows: 

In  the  first  place,  they  should  be  constructed  in  such  a  way  that 
the  momentum  which  has  been  acquired  by  the  liquids  and  solids  has 
been  destroyed.  This  has  been  accomplished  by  baffle  or  deflecting 
plates,  which  alter  or  reverse  the  flow  of  the  steam,  or  by  allowing  it 
to  expand  and  give  the  heavier  particles  time  to  fall  by  the  action  of 
gravity.  After  this  has  been  accomplished  it  is  important  to  prevent 
'the  separated  water  from  being  again  picked  up  and  carried  along  by 
the  purified  gases. 

Finally,  care  must  be  taken  that  an  ample  and  easy  passage  is  af- 
forded to  the  current  of  steam,  so  that  there  will  be  no  loss  of  energy 
from  friction. 

A  separator  should  be  placed  as  close  to  the  engine  cylinder  as  pos- 
sible. 

There  are  several  well  known  types  of  proved  merit,  made  for  both 
horizontal  and  vertical  pipes,  allowing  in  some  cases  to  placing  directly 
over  the  cylinder. 


Q.  37.  (1900-01.)  If  a  number  of  engines  are  receiving  their  steam 
supply  from  a  common  steam  main,  where  would  you  locate  the  sepa- 
rator? 

Ans.  37.  Should  there  be  a  number  of  engines  connected  to  the 
steam  main  and  only  one  separator  used,  it  would  be  placed  at  the 
last  engine.  This  answer  might  be  modified  in  certain  conditions  in 
which  the  pipe  was  not  in  the  same  plane.  The  proper  way  if  the 
engines  were  of  fair  size,  to  place  a  separator  at  each  engine.  The 
piping  off  to  the  engines  would  determine  the  position  of  the  separator 
in  the  line  of  the  main  steam  main. 

There  might  be  instances  where  the  separator  would  be  placed  near 
the  first  engine.  If  we  take  into  account  the  difference  in  the  specific 
gravity  of  water  and  steam,  the  momentum  obtained  by  the  entrained 
water  as  it  is  swept  along  with  the  steam  we  will  see  that  the  heavier 
particle  will  have  a  tendency  to  keep  in  a  right  line  direction  rather 
than  take  to  a  turn  in  direction,  thus  if  the  first  engine  was  piped  off 
from  a  side  outlet,  or  taken  from  the  top  of  the  pipe  there  would  be 
a  tendency  for  the  water  to  follow  the  straight  pipe.  Should  the  steam 
main  take  a  turn  at  right  angles  at  the  first  engine  then  the  probability 
is  that  the  separator  should  be  placed  near  first  engine. 


Q.  38.  (1900-01.)  In  erecting  a  steam  main  of  considerable  length, 
what  do  you  consider  the  best  practice  for  the  return  of  the  conden- 
sation? 

Ans.  38.  In  erecting  a  steam  main  of  considerable  length  a  large 
drum  or  separator  should  be  provided  near  the  engine,  for  trapping  the 
water  condensed  in  the  pipe. 

A  steam  drum  3  feet  in  diameter  and  15  feet  high  has  given  good 
results  in  separating  the  water  of  condensation  of  a  steam  pipe  10 
inches  in  diameter  and  800  feet  long. 

There  are  several  different  ways  of  conducting  the  water  of  conden- 
sation back  to  the  boiler.  The  steam  loop,  the  Holly  system,  pump  and 
receiver  system,  the  return  trap  system,  etc. 

Where  there  are  a  number  of  returns  of  the  same  pressure  coming 
back  bring  them  to  a  receiver  and  return  them  automatically  with 
pump.  This  system  gives  about  as  little  trouble  in  handling  as  any. 
Still  the  other  methods  have  their  advocates. 

114 


Q.  39.  (1900-01.)  Is  it  safe  practice  to  return  the  condensation 
from  the  steam  cylinders  of  pumps  or  engine's  cylinders  to  the  boiler? 
Explain  your  reasons,  "pro  and  con,"  setting  forth  dangers,  etc..  and 
preventives  for  eliminating  said  danger. 

Ans.  39.  In  localities  where  a  high  tariff  exists  on  water,  it  is 
deemed  advisable  to  save  and  re-use  all  of  the  drips  of  condensation. 
In  this  case,  care  should  be  taken  to  eliminate  from  the  exhaust  steam 
as  it  passes  from  the  steam  cylinders  all  oils  or  greases.  It  is  also 
advisable  to  use  in  such  cases  a  high  grade  of  mineral  oil  for  cylinder 
lubrication. 

In  discussing  the  reasons  "pro  and  con,"  they  could  be  made  the 
subject  of  an  extended  paper,  but  while  it  is  easily  recognized  that  it 
is  better  not  to  return  them  to  the  boiler,  economy  says  in  this  instance 
that  we  must. 

A  discussion  can  easily  arise  as  to  the  merits  and  demerits  of  open 
vs.  closed  heaters,  etc. 

In  which  ever  way  it  is  done  constant  care  should  be  taken  that  all 
apparatus  is  kept  in  proper  condition  and  the  water  returned  as  free 
from  oil  and  foreign  elements  as  possible. 

It  is  good  practice  to  feed  in  some  service  water  at  all  times  to  make 
up  what  little  deficiency  is  lost  by  evaporation  or  leaks. 


Q.  40.  (1900-01.)  In  connecting  drips  from  a  number  of  engines  into 
a  common  drip  line,  what  safeguard  should  be  applied? 

Ans.  40.  Several  drip  lines  (all  under  the  same  pressure)  connected 
to  a  main  drip  line,  should  have  a  check  valve  to  prevent  the  steam  or 
condensation  from  backing  up  from  any  individual  line. 


Q.  41.   (1900-01.)     What  are  the  advantages  of  high  piston  speed? 
What  kind  of  service  requires  high  rotative  speed?    Why? 
Why  should  high  speed  engines  run  under  high  pressure? 

Ans.  41.  The  advantages  of  high  piston-speed  over  slow-speed  en- 
gines may  be  briefly  stated  as  follows: 

.b'irst — For  a  given  steam  pressure  and  cut-off  the  power  of  an  engine 
varies  directly  as  its  speed.  There  are  four  factors  which  determine 
the  power  of  an  engine,  viz: 

(a)  The  mean  effective  pressure  on  the  piston. 

(b)  The  length  of  the  stroke. 

(c)  The  area  of  the  piston. 

(d)  The  speed. 

In  other  words,  an  engine  of  a  given  diameter  and  length  of  stroke, 
acting  under  a  given  mean  effective  pressure,  will  develop  power  in 
proportion  to  its  speed,  and  if  the  speed  is  doubled  its  power  will  also 
be  doubled,  and  so  to  obtain  a  given  power  under  a  given  mean  effec- 
tive pressure  we  need  make  an  engine  only  half  as  large  if  we  double 
its  speed. 

This  constitutes  the  first  argument  for  high  speeds,  economy  both  in 
first  cost  and  in  space. 

Secondly — Where  power  is  transmitted  to  shafting  or  directly  to 
machine  running  at  a  much  higher  speed,  it  can  be  performed  more 
efficiently  if  the  ratio  of  the  speeds  is  not  too  great. 

xn  many  cases  when  the  power  is  transmitted  by  belting  the  ratio 
of  speed  is  too  great  to  admit  of  transmission  in  a  single  step,  since  the 
arc  of  contact  on  the  driven  pulley  would  be  too  small  to  prevent  slip- 
page of  the  belt.  In  such  cases  it  is  common  to  use  an  intermediate 
shaft  which  performs  no  other  duty  than  to  make  the  reduction  more 
gradual,  and  thus  insure  a  satisfactory  running  of  the  belt, 

"5 


By  increasing  the  speed  of  the  engine  this  is  done  away  with  in 
many  cases. 

In  the  case  of  dynamo  machines  it  is  now  common  practice  to  couple 
the  shaft  of  the  engines  and  dynamo  direct  without  any  belting  what- 
ever. 

In  spite  of  all  that  may  be  said  against  this  practice,  it  cannot  be 
denied  that  it  saves  a  very  considerable  amount  of  valuable  space. 

Thirdly — It  is  claimed  for  high-speed  steam  engines  that  tue  econ- 
omy in  the  use  of  steam  far  exceeds  that  of  the  older  forms  of  slow- 
speed  engines. 

There  is  no  doubt  a  great  deal  of  justice  in  this  claim,  because  one 
of  the  main  losses  in  the  engine,  viz.,  the  cooling  of  the  cylinder  walls 
and  passages  during  exhaust  and  re-evaporation,  is  greatly  reduced. 
The  disadvantage  of  admitting  steam  by  the  same  passage  through 
which  it  exhausts  is  clearly  demonstrated  in  the  increased  economy  of 
the  Corliss  type  of  engine,  where  this  is  not  the  case.  As  far  as  re- 
evaporation  is  concerned,  it  is  obvious  that  the  cylinder  walls  are  chilled 
very  considerably  during  this  process,  and  the  more  steam  passed 
through  the  engine  in  a  given  space  of  time  the  less  will  be  the  re- 
evaporation.  Hence  the  increased  economy  of  the  high-speed  engine. 

It  is  a  fact  that  the  uniformity  and  smoothness  of  running  of  the 
high-speed  engines  make  them  particularly  adapted  for  dynamo  run- 
ning. This  uniformity  of  running  is  in  part  due  to  the  fact  that  the 
influence  of  the  fly-wheel  is  greatly  enhanced  and  partly  to  the  steady- 
ing influence  of  the  reciprocating  parts.  It  is  a  well-known  fact  that 
the  steadying  influence  of  a  fly-wheel  is  proportional  to  the  square  of 
its  speed. 

For  instance,  if  one  engine  runs  twice  as  fast  as  another  of  same 
design  and  the  same  weight  of  fly-wheel,  it  will  run  four  times  as 
steady.  Now,  in  every  steam  engine  the  pressure  is  a  maximum  at  the 
beginning  of  the  stroke,  decreasing  as  the  stroke  advances  until  the 
end,  when  it  is  a  minimum;  hence  it  is  evident  that  the  action  of  the 
reciprocating  parts,  which  is  to  absorb  or  store  a  portion  of  the  pressure 
during  the  first  half  of  the  stroke  and  restore  it  during  the  second  half, 
has  the.  effect  of  tending  to  keep  the  pressure  on  the  crank  uniform 
during  the  entire  stroke,  or,  in  other  words,  to  steady  the  running  of 
tne  engine. 

High-speed  engines  must  be  high  pressure  engines — that  is,  the 
initial  steam  pressure  must  be  sufficient  always  to  set  the  reciprocating 
parts  in  motion,  and  the  pressure  should  never  be  reduced  by  the 
means  of  the  throttle-valve,  otherwise  the  engine  will  be  subjected  to 
strains  which  will  impair  its  life. 

The  steam  passages  and  parts  should  be  of  ample  size,  the  higher  the 
speed  the  larger  the  passages,  otherwise  the  pressure  will  be  reduced 
before  entering  the  cylinder. 

The  steam  should  be  cut  off  early,  and  a  moderate  amount  of  com- 
pression to  provide  cushioning  the  reciprocating  parts  as  they  come  to 
rest  at  the  end  of  each  stroke  should  be  provided  for  by  an  early  closing 
of  the  exhaust  valve. 


Q.  42.  (1900-01.)     What  are  rotary  engines? 

Why  are  they  not  more  extensively  used? 

What  are  the  principal  sources  of  loss  in  steam  engines? 

What  is  the  principal  source  of  waste  in  the  engine  proper? 

How  may  it  be  reduced? 

Ans.  42.  In  the  rotary  engine  there  is  no  reciprocating  motion  what- 
ever, the  force  of  the  steam  being  used  to  produce  at  once  a  motion  of 
rotation.  Rotary  steam  engines  other  than  steam  turbines  have  been 
invented  by  the  thousands,  but  no  one  has  attained  a  commercial  suc- 

116 


cess.    The  possible  advantages,  such  as  speed  and  the  saving  of  space, 
to  be  gained  by  the  rotary  engine  are  overbalanced  by  its  waste  of 


In  designing  and  constructing  steam  engines  for  additional  econ- 
omy in  the  use  of  fuel,  and  yet  with  all  that  has  been  done  in  this 
direction  since  the  time  of  James  Watt,  the  steam  engine  is  still  a 
very  wasteful  machine,  and  only  a  small  percentage  of  the  energy  con- 
tained in  the  fuel  is  actually  converted  by  the  steam  engine  into  useful 
work. 

There  are  certain  losses  of  energy  which  it  has  hitherto  been  im- 
possible to  avoid,  and  so  long  as  these  must  be  incurred  so  long  will 
the  process  of  transforming  the  latent  energy  of  the  coal  into  mechani- 
cal work  by  the  means  of  the  steam  engine  be  a  wasteful  one.  It  has 
been  demonstrated  that  in  the  use  of  high-speed  non-condensing  type 
that  of  the  total  energy  contained  in  the  fuel  that  only  from  5  per  cent 
to  10  per  cent  is  available  at  the  shaft  of  the  engine.  A  large  part  of 
this  loss  of  energy  occurs  in  the  steam  generator,  due  in  part  to  the 
high  temperature  of  the  gases  escaping  into  the  chimney,  which  is 
necessary  for  the  production  of  the  draught,  and  partly  to  the  excess  of 
air  over  and  above  that  necessary  for  the  complete  combustion  of  the 
fuel.  Additionally  there  are  losses  due  to  radiation,  to  leakage  and  to 
the  heat  carried  away  in  the  ashes. 

The  principal  source  of  waste  in  the  engine  proper  is  cylinder  con- 
densation and  re-evaporation.  The  condensation  of  steam  in  the  cyl- 
inder is  greater,  the  greater  the  surface  exposed  for  a  given  weight 
of  steam  passing  through  the  engine,  consequently  it  is  not  as  great 
proportionately  in  large  engines  as  in  the  smaller  ones. 

Obviously,  also,  it  is  proportional  to  the  range  of  temperature  to 
which  the  cylinder  is  exposed,  and  it  is  for  this  reason  that  multiple 
expansion  engines  are  more  economical. 

As  for  the  condensation  in  the  steam  passages,  it  has  been  found 
that  this  could  be  almost  entirely  eliminated  by  closing  the  exhaust 
before  the  end  of  the  stroke,  so  as  to  compress  the  steam  and  thus 
raise  the  temperature  of  the  walls  and  passages,  or  more  effectually 
by  providing  separate  passages  for  admission  and  exhaust. 

Losses  in  the  engine  proper  by  the  presence  of  moisture  in  the 
steam  cylinder,  which  is  carried  into  the  cylinder  with  the  steam  from 
the  boilers  or  is  produced  by  the  condensation  on  the  cylinder  walls 
or  in  the  steam  passages,  is  re-evaporated  during  the  expansion  and 
exhaust  periods,  and  re-evaporating,  abstracts  heat  from  the  steam. 
Very  little  of  this  heat  which  is  used  in  this  way  is  returned  to  the 
engine  as  useful  work,  and  consequently  the  presence  of  moisture  may 
be  said  to  rob  the  engine  of  an  amount  of  useful  energy  proportional  to 
the  heat  required  for  its  evaporation. 

It  is,  therefore,  a  matter  of  economy  to  prevent  the  entrance  of 
water  to  the  cylinder  or  its  formation  by  condensation  on  the  walls 
of  the  cylinder  and  in  the  steam  passages. 


Q.  44.  (1900-01.)  Which  is  the  most  economical  type  of  engine,  and 
why? 

What  is  the  cause  of  cylinders  wearing  unevenly? 

How  can  it  be  avoided? 

What  is  meant  by  the  term  "clearance"?  ,  .  j 

Why  is  it  a  necessity  anu  how  can  it  be  determined  in  a  given 
engine? 

Ans.  44.  The  four-valve  and  the  Corliss  type  of  engines  are  more 
economical  than  those  in  which  the  same  passages  are  used  for  both 
admission  aini  exhaust. 

The  losses  of  heat  due  to  radiation  may  be  effectually  prevented  by 
surrounding  the  cylinder  with  non-conducting  material. 

117 


The  Steam  jacket  is  also  frequently  used  and  is  effective  in  reducing 
the  losses  due  to  initial  condensation. 

It  is  a  general  impression  among  engineers  that  the  cylinders  of 
very  large  horizontal  engines  are  more  liable  to  wear  oblong  than 
those  of  vertical  engines  of  the  same  bore;  but  experience  and  obser- 
vation have  proved  this  to  be  a  mistaken  idea. 

The  trouble  is  frequently  due  to  imperfect  alignment,  and  it  is  dif- 
ficult to  imagine  why  an  engine  piston  should  exert  any  pressure 
against  the  cylinder  walls,  other  than  that  due  to  gravity,  when  all  the 
reciprocating  parts  are  perfectly  true  with  the  center  line  of  the 
engine  and  with  each  other. 

Care  should  be  taken  in  packing  the  rod  that  there  should  be  no 
inequality  in  the  packing,  as  any  material  inequality  may  throw  the 
engine  out  of  alignment. 

The  wear  which  would  occur  on  the  bottom  of  large  cylinders  of 
large  horizontal  engines  on  account  of  the  weight  of  the  piston  is 
frequently  avoided  by  extending  the  rod  through  a  stuffing-box  in  the 
outer  cylinder  head. 

The  term  clearance  is  understood  to  mean  the  unoccupied  space  be- 
tween the  piston  and  cylinder-heads  when  the  crank  is  on  the  dead 
center;  but  it  also  applies  to  the  space  between  the  cylinder  and  the 
face  of  the  valves. 

The  amount  of  clearance  of  any  engine  affects  its  economy;  that  is, 
if  the  clearance  is  small,  the  engine  will  be  more  economical  than  if 
large;  for  obvious  reasons  a  certain  amount  of  clearance  is  a  necessity 
to  prevent  the  piston  from  striking  the  heads  of  the  cylinder,  due  to 
the  wearing  of  boxes  and  pins  or  an  unequal  adjustment  of  the  different 
reciprocating  parts. 

The  most  accurate  method  of  ascertaining  the  exact  amount  of  clear- 
ance is  to  weigh  up  a  certain  amount  of  water,  place  the  engine  on  the 
dead  center,  and  then  from  this  weighed  or  measured  water  fill  the 
clearance  space  up  to  the  face  of  the  steam  valve;  reweighing  the 
water  remaining  and  subtracting  from  the  first  weight  will  give  the 
number  of  pounds  that  are  required  to  fill  the  clearance  space,  which 
can  be  reduced  to  cubic  inches,  and  in  comparison  with  the  cubic  con- 
tents of  the  cylinder,  the  percentage  is  easily  arrived  at. 


Q.  45.  (1900-01.)  What  consideration  determines  the  length  of  con- 
necting rods? 

How  long  are  they  usually  made? 

What  is  the  function  of  the  cross-head? 

What  are  three  different  forms  of  cranks? 

Where  would  they  be  used? 

Explain  the  difference  of  the  terms  "throwing  over"  and  "throwing 
under"? 

Which  way  is  advisable? 

What  are  the  disadvantages  of  center  cranks? 

Of  what  material  should  they  be  constructed? 

Ans.  45.  The  length  of  stroke  is  one  of  the  principal  considerations 
that  govern  the  length  of  the  connecting  rod.  The  connecting  rod,  like 
the  piston  rod,  is  subjected  to  both  a  tensile  and  compressive  stress; 
undue  stress,  due  to  accidents,  etc.,  must  be  taken  into  consideration. 
Long  connecting  rods  have  many  advantages,  but  the  longer  they  are 
the  greater  must  be  their  thickness,  and  they  are  not  as  economical 
in  the  use  of  material  as  the  short  connecting  rods.  For  long-stroke 
engines  they  are  generally  made  from  two  to  four  times  the  length  of 
H>6  stroke.  The  usual  length  for  high-speed  engines  is  five  times  the 
length  of  the  crank. 

118 


The  cross-head  is  that  part  of  the  engine  which,  moving  between 
guides,  preserves  the  rectilinear  motion  of  the  piston  rod  and  at  the 
same  time  supplies  a  bearing,  called  a  wrist-pin,  for  the  rocking  motion 
of  one  end  of  the  connecting  rod. 

A  crank  is  a  simple  lever,  at  one  end  of  which  acts  the  steam  pres- 
sure which  is  transmitted  to  it  by  the  reciprocating  parts,  while  the 
other  is  secured  to  the  shaft  of  the  engine.  It  is  the  final  link  in  the 
transformation  of  reciprocating  into  rotary  motion.  The  three  differ- 
ent forms  of  crank  are  the  side  crank,  disc  crank,  and  the  center  crank. 
The  center  cranks  are  used  wnen  the  shaft  extends  on  either  side  of 
the  crank.  They  are  usually  forged  in  one  piece  with  the  shaft,  but 
weaken  it  somewhat. 

The  side  and  disc  cranks  can  be  used  in  any  case  where  the  connect- 
ing rod  is  on  the  side  of  the  engine.  The  advantage  of  the  disc  over 
the  crank  is  that  it  affords  better  facilities  for  balancing. 

An  engine  throws  over  when  the  crank-pin  traverses  the  upper  por- 
tion of  its  travel  while  the  piston  is  moving  towards  the  main  shaft  and 
throws  under  when  the  crank-pin  traverses  the  lower  portion  of  its 
travel  while  the  piston  is  moving  toward  the  main  shaft. 

In  the  first  case  the  stress  on  the  guides  is  downward,  which  is 
preferable;  in  the  second  case  the  stress  is  upward.  Engines  are  usu- 
ally built  to  throw  over,  and  it  is  only  advisable  to  throw  under  in 
such  cases  where  the  transmission  is  such  that  the  tight  side  of  the 
belt  is  on  the  top  of  the  pulley,  which  is  never  advisable. 


Q.  46.   (1900-01.)     What  is  an  eccentric? 
In  what  respect  does  it  differ  from  a  crank? 
When  is  it  used  in  preference  to  a  crank?    Why? 
Explain  the  throw  of  the  eccentric. 

Ans.  46.  An  eccentric  is  substantially  a  crank,  with  its  pin  enlarged 
in  diameter  so  as  to  enclose  the  shaft  on  which  it  is  placed  within  its 
periphery-  It  gives  exactly  the  same  motion  that  would  be  obtained 
from  an  ordinary  crank  of  equal  throw. 

Eccentrics  are  generally  used  for  converting  rotary  into  reciprocat- 
ing motions,  while  cranks  are  used  for  the  opposite  purpose,  although 
the  latter  can  accomplish  both  results. 

The  principal  reason  why  eccentrics  are  used  instead  of  cranks  to 
actuate  the  valve  gear  is  because  the  motion  must  generally  be  taken  off 
at  some  point  near  the  middle  of  the  shaft,  hence  a  center-crank  would 
be  required,  which  would  weaken  the  shaft.  The  distance  between  the 
center  of  the  eccentric  sheave  and  the  center  of  the  shaft  is  called  the 
throw  of  the  eccentric  or  eccentricity. 


Q.  47.  (1900-01.)  Calculate  the  diameter  of  a  shaft  that  will  safely 
transmit  1,000  HP.  at  100  R.  P.  M.,  considering  torsional  stresses  only. 

If  the  transverse  stresses  were  taken  into  consideration,  how  would 
you  proceed  to  find  the  true  diameter? 

Ans.  47. 
Rule— 

(a)  If  steel,  multiply  the  H.  P.  to  be  transmitted  by  75  and  divide 
the  product  by  the  number  of  revolutions  per  minute.     Extract  the  cube 
root  of  the  results,  which  will  be  the  required  diameter  of  the  shaft  in 
inches. 

(b)  If  wrought  iron,  multiply  the  H.  P.  to  be  transmitted  by  100, 
then  proceed  as  above. 

Example — Use  formula  for  steel  shaft. 

119 


1000  X  75 

=  SV  750  =  9.0856  inches  =  dia.  of  shaft. 


100 

The  usual  method  to  determine  the  proper  diameter  for  the  crank 
shaft  is  to  calculate  it  according  to  the  above  rule,  then  consider  it  as  a 
beam  carrying  a  load  equal  to  the  total  maximum  pressure  of  the  steam 
on  the  piston,  and  determine  what  diameter  would  be  necessary  to 
safely  carry  the  load.  The  greater  of  the  two  results  will  be  the  proper 
diameter  for  the  shaft. _____ 

Q.  48.   (1900-01.)     What  are  the  functions  of  a  "fly-wheel"? 

Where  should  the  bulk  of  the  metal  be  concentrated? 

Should  it  be  evenly  balanced? 

How  would  you  proceed  to  find  the  safe  weight  of  a  fly-wheel  for 
any  size  or  style  of  engine? 

Suppose  an  engine  is  12  "  X  24  ",  making  140  R.P.M.,  diameter  of 
wheel  6  feet;  what  would  be  the  proper  weight  of  the  wheel? 

Ans.  48.  The  function  of  a  fly-wheel  is  to  equalize  the  motion  when- 
ever either  the  power  communicated  or  the  resistance  to  be  overcome  is 
variable.  In  one  case  where  the  fly-wheel  is  used  to  overcome  a  varia- 
ble resistance,  it  may  be  considered  a  conservator  of  power. 

In  the  other  case  the  fly-wheel  may  be  said  to  be  a  distributer  of 
power. 

The  fly-wheel,  as  before  stated,  is  a  regulator  and  a  reservoir,  and 
not  a  creator  of  motion.  As  regularity  of  motion  is  of  much  greater 
importance  in  some  cases  than  in  others,  the  weight  and  diameter  of  the 
fly-wheel  must  depend  upon  the  work  and  character  of  the  machinery 
it  is  intended  to  drive;  so  that  in  proportioning  a  fly-wheel  to  a  given 
engine,  attention  must  be  paid  to  many  particular  circumstances  rather 
than  to  any  given  rule. 

The  effectiveness  of  the  fly-wheel  in  steadying  the  motion  of  the 
engine  depends  upon  the  distance  of  the  metal  from  the  center.  For 
this  reason  the  material  of  which  the  fly-wheel  is  composed  should  be 
concentrated  as  much  as  possible  in  the  rim.  The  steadying  action  also 
varies  as  the  square  of  the  speed  of  the  rim.  Hence  within  certain 
limits  increasing  the  diameter  saves  weight. 

The  speed  of  the  rim  is  rarely  above  80  feet  per  second,  and  if  car- 
ried beyond  200  feet  per  second  the  strains  produced  by  centrifugal 
force  would  probably  be  sufficient  to  rupture  the  wheel. 

Great  care  should  be  taken  in  erecting  fly-wheels  to  see  that  they 
are  perfectly  balanced— that  is,  that  the  center  of  gravity  of  the  wheel 
coincides  with  the  center  of  the  shaft. 

Rule— For  finding  the  weight  of  a  fly-wheel  having  the  size  of  the 
cylinder,  the  diameter  of  the  wheel  and  the  revolutions  per  minute 
given —  , 

First — Multiply  area  of  the  piston  by  the  length  of  stroke  in  feet. 
Multiply  this  product  by  constant,  12,000,000. 

Second— Square  the  number  of  revolutions.  Multiply  this  by  the 
square  of  the  diameter  of  the  wheel  in  feet. 

Divide  the  first  result  by  the  second  and  the  quotient  is  the  proper 
weight  of  the  fly-wheel  in  pounds. 

Example — As  per  question — 

12  in.  dia.  =  113  in.  area. 

First  113  X  2  X  12,000,000  =  2,712,000,000. 

Second  1402  X  62  =  705,600. 

2,712,000,000  -^  705,600  =  3,857  Ibs.— The  weight  of  the  fly-wheel 


Q.  49.  (1900-01.)  Explain  the  terms:  "-Valve  gear"  "releasing  valve 
gear,"  "automatic  cut-off,"  "positive  cut-off,"  "riding  cut-off"  and  "re- 
versing gear." 


Ans.  49.  The  term  valve-gear  embraces  all  intermediate  connections 
between  the  eccentric  on  the  driving-shaft  and  the  valves,  and  is  appli- 
cable to  all  mechanical  arrangements  employed  for  working  the  valves 
of  steam  engines. 

Releasing  valve-gear  is  an  arrangement  in  which  the  valve  is  lib- 
erated from  the  control  of  its  moving  agent,  and  allowed  to  close  in 
obedience  to  the  action  of  a  spring,  weight  or  other  force  independent 
of  that  which  opened  it. 

An  automatic  cut-off  valve-gear  is  one  in  which  the  movement  of  the 
cut-off  vaive  is  so  controlled  by  the  governor  as  to  cut  off  the  steam  as 
early  or  as  late  in  the  stroke  as  may  be  required,  to  maintain  the  desired 
uniformity  of  speed,  under  variations  of  load  and  pressure. 

Positive  cut-off  is  an  arrangement  of  valve-gear  by  which  the  expan- 
sion of  steam  is  effected  by  what  is  known  as  lap  on  the  valve,  the 
steam  being  cut  off  at  the  same  point  in  each  stroke,  independent  of 
load  or  pressure. 

Riding  cut-off  is  a  term  applied  to  cut-off  valves  which  ride  on  the 
back  of  the  main  steam  valve. 

A  "reversing"  valve-gear  is  an  arrangement  employed  for  reversing 
the  motion  of  engines.  It  is  effected  in  different  ways:  in  some  cases 
with  a  single  eccentric,  while  in  others  with  two  eccentrics,  as  in  the 
case  of  the  link;  and  in  others  still,  by  the  means  of  a  loose  eccentric 
which  revolves  on  the  shaft,  but  is  prevented  from  making  a  complete 
revolution  by  two  stops  so  placed  that  one  arrests  it  in  the  proper  posi- 
tion for  the  forward,  and  the  other  for  the  backward  motion.  This  last 
arrangement  is  peculiarly  adapted  to  tug-boats  and  ferries,  owing  to  the 
ease  and  quickness  with  which  the  engine  can  be  reversed. 


Q.  50.  (1900-01.)  What  are  "relief  valves,"  "balance  valves,"  "ro- 
tary valves"  and  "gridiron  valves"? 

Describe  the  plain  slide  valve  and  its  action. 

Why  are  plain  slide  valves  not  used  in  large  engines  and  high  pres- 
sures? 

Ans.  50.  Relief  valves  are  used  on  the  cylinders  of  large  engines 
to  prevent  fracture  of  the  cylinder-head  and  cylinder,  in  consequence 
of  an  accumulation  of  water  in  the  latter.  They  are  also  used  in  many 
places  for  a  relief  from  overpressure. 

Balance-valves  are  arrangements  by  which  the  weight  on  the  back 
oi  slide-valves,  induced  by  the  pressure  of  the  steam,  is  relieved  by  the 
action  of  the  steam  in  the  steam-chest. 

Rotary-valves  is  a  term  applied  to  any  valve  that  describes  a  revolu- 
tion in  working. 

Semi  rotary-valves  is  a  term  applied  to  all  valves  that  have  a  vibra- 
tory or  rocking  motion,  similar  to  the  Corliss. 

Gridiron-valves  are  a  modification  of  the  slide-valve,  containing  a 
number  of  openings  for  the  steam,  by  which  means  its  travel  and  fric- 
tion are  materially  diminished.  Multi-ported  valves. 

The  function  of  the  common  slide-  valve  is  to  admit  steam  to  the 
piston  at  such  times  when  its  force  can  be  usefully  expended  in  pro- 
pelling it,  and  to  release  it  when  its  pressure  in  the  cylinder  is.  no 
longer  required. 

Owing  to  the  amount  of  power  expended  in  moving  the  valve  in 
large  engines  and  high  pressures,  the  plain  slide-valve  is  but  little  used. 


Q.  51.  (1900-01.)  Define  the  terms  "admission,"  "exhaust,"  "cut-off," 
"expansion,"  "compression,"  "angle  of  advance,"  "travel,"  "over  travel," 
"inside  lap,"  "outside  lap,"  "steam  lead,"  "exhaust  lead,"  and  "negative 
lead." 

T2I 


Ans.  51.  Admission:— The  period  during  which  the  steam  passages 
are  open  and  steam  is  admitted  behind  the  piston. 

Exhaust:— The  period  during  which  the  exhaust  passages  are  open 
and  the  steam  is  exhausted  from  the  cylinder. 

Cut-off:— The  point  in  the  stroke  at  which  the  steam-valve  closes. 

Expansion: — The  period  during  which  the  steam  expands  in  the 
cylinder,  beginning  at  the  point  of  cut-off  and  continuing  until  the 
steam  is  released  or  the  exhaust  port  is  opened. 

Compression:— The  period  during  which  the  steam  is  compressed, 
which  is  from  th«  time  the  exhaust  port  closes  until  the  end  of  the 

Angular  advance:— The  angle  which  would  be  formed  by  the  eccen- 
tric when  in  its  actual  position  with  the  position  of  the  eccentric  cor- 
responding to  the  central  position  of  the  valve,  the  crank  being  on  its 
dead  point. 

Travel  of  the  valve: — The  total  distance  which  the  valve  moves  in 
one  direction. 

If  the  valve  rod  was  of  infinite  length,  this  would  be  equal  to  twice 
the  throw  of  the  eccentric. 

Overtravel: — Is  the  distance  traveled  by  the  valve  over  and  above 
tnat  necessary  to  fully  open  the  steam  port 

Lap  on  the  valve:— The  term  lap  on  the  valve  denotes  the  amount 
the  edges  of  the  valve  extend  over  the  ports  when  the  valve  is  in  the 
center  of  its  travel.  The  object  of  lap  is  to  secure  the  benefit  to  be  de- 
rived from  the  working  of  steam  expansively. 

Lap  on  the  steam  side  of  the  valve  is  termed  outside  lap,  while  lap 
on  the  exhaust  side  is  termed  inside  lap. 

Steam  lead:— Is  the  amount  the  port  is  open  at  the  beginning  of  thp 
stroke.  The  object  of  lead  is  to  enable  the  steam  to  act  as  a  cushion 
against  the  piston  before  it  arrives  at  the  end  of  the  stroke,  to  cause 
it  to  reverse  its  motion  easily  and  also  to  supply  steam  of  full  pressure 
to  the  piston  the  instant  it  passes  the  dead  center.  Generally  the  high- 
er the  speed  and  the  more  irregular  the  work,  the  more  lead  will  be 
required  for  any  engine. 

Lead  varies  in  different  engines  from  1-32  to  3-16  of  an  inch.  Some 
valves  have  no  lead  at  all,  others  less  than  none,  or  what  is  termed 
negative  lead.  Lead  on  the  exhaust  end  means  the  amount  of  open- 
ing the  valve  has  on  the  end  from  which  the  steam  is  escaping.  The 
name  applies  alternately  to  each  end  of  the  cylinder. 


Q.  52.  (1900-01.)  Given  the  throw  of  the  eccentric,  angle  of  advance 
and  inside  and  outside  lap.  show  how,  by  the  "Zeuner  diagram,"  the 
distribution  of  steam  may  be  studied. 

Ans.  52.  Draw  a  line  OX  to  represent  the  crank  at  the  beginning 
of  the  stroke,  and  with  this  as  a  radius  draw  the  crank  circle  XX1,  X\ 
*.s,  X4.  Suppose  the  crank  to  turn  in  the  direction  of  the  arrow. 

Through  the  point  O  draw  the  line  RR1,  making  the  angle  R'OY1 
equal  to  the  angle  of  advance,  and  lay  off  the  distances  OR  and  OR1 
equal  to  the  eccentricity  or  throw  of  the  eccentric.  On  the  lines  OR 
and  OR1  as  diameters  draw  the  two  circles  ORCD  and  OER'F. 

With  0  as  a  center  and  a  radius  OA  equal  to  the  outside  or  steam 
lap  draw  a  circle  ACD,  and  similarly  with  a  radius  OB  equal  to  the 
inside  or  exhaust  lap  draw  a  circle  BEF.  Through  the  point  O  and  the 
intersections  C,  D,  E  and  F  draw  the  lines  OX1;  OX2,  OXs  and  OX4. 

We  are  now  able  to  take  from  the  diagram  all  of  the  data  necessary 
for  a  complete  understanding  of  the  distribution  of  steam  in  the  cylin- 
der: 

OXi  is  the  position  of  the  crank  when  admission  of  steam  takes 
place. 


OXa  is  thfe  position  of  the  crank  when  cut-off  takes  place.  Hence 
Xt  OX2  is  the  angle  traversed  by  the  crank  during  the  period  of  ad- 
mission. 

OXs  is  the  position  of  crank  when  exhaust  opens. 

OXi  is  the  position  of  crank  when  exhaust  closes,  hence  XsOX4  is 
the  angle  traversed  by  the  crank  during  the  period  of  exhaust  and 

X^OXi  is  the  angle  traversed  by  the  crank  during  the  period  of  com- 
pression. 

The  distances  from  the  intersection  of  the  circles  R  and  Ri  with  the 
lines  OX3  OX,,  etc.,  representing  the  crank  in  its  different  positions  to 
the  center,  represent  the  travel  of  the  valve  corresponding  to  those 
positions  of  the  crank. 

The  circle  R  represents  the  forward  and  the  circle  R1  the  return 
stroke,  hence  OK  is  the  distance  the  valve  has  traveled  from  its  central 
position  at  the  beginning  of  the  stroke. 

OK1,  the  same  for  return  stroke. 

OA  is  the  outside  or  steam  lap,  hence  AK  is  the  distance  the  steam 
port  is  open  at  the  beginning  of  the  stroke  or  steam  lead. 

OR  is  the  full  travel  of  the  valve. 

OB  is  the  inside  or  exhaust  lap,  hence  BK  is  the  distance  the  ex- 
haust port  is  open  at  the  beginning  of  the  stroke  or  the  exhaust  lead. 


THE   ZEUNER  DIAGRAM. 

At  the  points  C  and  D  the  travel  of  the  valve  is  just  equal  to  the 
outside  lap;  hence  in  these  positions  of  the  crank  the  steam  port  opens 
and  closes  respectively;  similarly  at  the  points  E  and  F  the  travel  is 
just  equal  to  the  exhaust  lap;  hence  in  these  positions  of  the  crank 
the  exhaust  port  opens  and  closes  respectively. 

If  we  lay  down  from  the  point  A  at  a  distance  AH,  equal  to  the 
width  of  the  port,  and  with  O  as  a  center  and  a  radius  OH,  draw  an 
arc  cutting  the  line  OR,  at  J. 

JR  is  the  distance  the  valve  travels  more  than  enough  to  fully  open 
the  port,  or  the  over-travel.  Similarly  if  we  lay  off  from  B  the  dis- 

123 


tance  BL  equal  to  the  width  of  the  port,  and  from  the  center  O  and  a 
radius  equal  to  OL  draw  an  arc  cutting  OR  at  M,  MR  is  the  distance  the 
valve  travels  more  than  enough  to  fully  open  the  port  for  exhaust. 

It  will  thus  be  seen  that  by  a  careful  study  of  the  diagram  all  in- 
formation necessary  for  the  proper  design  and  setting  Of  the  valve  gear 
may  readily  be  had.  For  example,  in  the  above  diagram  the  cut-off 
takes  place  a  little  later  than  %  stroke.  It  is  evident  that  if  it  is  de- 
sired to  have  the  cut-off  take  place  earlier,  say  y2  stroke,  it  will  be 
necessary  for  the  outside  lap  circle  ACD  to  intersect  the  valve  circle  R 
in  the  line  YY1.  This  may  be  accomplished  by  increasing  the  outside 
lap,  by  reducing  the  eccentricity,  or  by  changing  the  angle  of  advance. 
However,  any  one  of  these  changes  would  also  effect  the  entire  distri- 
bution, and  it  would  probably  be  necessary  to  lay  down  several  dia- 
grams before  the  most  advantageous  dimensions  could  be  obtained. 


Q.  53.  (1900-01.)  Explain  how  the  lap  and  lead  may  be  determined 
without  removing  the  cover  to  steam  chest. 

Describe  the  piston  valve  and  state  its  advantages  and  disadvantages 
as  compared  to  a  plain  side  valve. 

Ans.  53.  Open  the  cylinder  drain  cocks  and  disconnect  them  from 
the  drain-pipe,  so  that  the  steam  may  be  seen  and  heard  to  issue  from 
them.  Or,  open  the  holes  made  for  the  indicator,  if  there  are  any; 
then  let  in  a  little  steam  and  turn  the  engine  over  by  hand,  and  note 
the  commencement  and  cessation  of  the  flow  of  the  steam,  just  when 
the  steam  is  admitted  and  cut  off.  The  point  of  cut-off  can  be  most 
accurately  ascertained  by  turning  the  engine  backwards.  The  steam 
in  this  case  will  commence  blowing  at  the  same  point  of  the  stroke  at 
which  it  would  cease  blowing  when  turning  it  forward;  and,  owing 
to  the  elasticity  of  the  steam,  the  commencement  of  the  issue  is  always 
more  clearly  defined  than  the  cessation  when  the  issuing  orifice  is  small. 
For  the  same  reason,  the  point  of  admission  can  be  most  accurately 
located  by  turning  the  engine  forward. 

To  determine  the  lead,  having  found  the  point  of  admission,  make 
a  mark  on  the  valve-stem  at  a  known  distance  from  some  fixed  point, 
and  another  after  the  pin  has  reached  the  dead  center;  this  will  give 
the  lead.  If  the  admission  forward  takes  place  when  the  crank-pin  is 
exactly  on  the  dead  center,  there  is  no  lead.  Having  obtained  the  lead 
and  cut-off  for  both  ends,  the  travel  and  length  of  the  connection  being 
known,  a  diagram  may  be  constructed  similar  to  the  one  in  the  previous 
question,  which  will  give  the  lap  and  port  openings. 

To  obviate  the  extreme  amount  of  friction  of  the  common  slide-valve, 
especially  those  of  large  size,  where  high  pressure  is  used,  the  piston- 
valve  has  been  designed  to  overcome  this  difficulty,  it  being  a  per- 
fectly balanced  valve,  and  the  only  pressure  on  the  seat  is  due  simply 
to  the  weight  of  ihe  valve.  As  usually  constructed,  the  valve-chest  is 
bushed,  the  bushing  being  accurately  turned  to  form  the  valve-seat, 
and  the  valve  is  made  tight  by  the  use  of  piston  rings,  the  same  as 
steam  pistons. 

In  the  Armington  and  Sims  type  it  will  be  seen  that  the  steam 
enters  the  cylinder  around  the  inside  edge  of  the  valve,  and  also 
through  additional  passage  cut  in  the  valve.  In  this  way  the  effective 
opening  of  the  port  for  a  given  travel  is  greatly -increased.  The  ex- 
haust takes  place  around  the  outside  of  the  valve,  so  that  in  reality 
the  inside  is  the  steam  lap  and  the  outside  is  the  exhaust  lap  opposite 
as  in  the  case  of  the  ordinary  D-slide  valve. 

The  principal  objection  to  piston-valves  is  that  the  seat  wears  un- 
evenly— that  is;  at  the  bottom  only,  and  thus  become  leaky.  However, 
the  seat  may  be  readily  removed  and  replaced  by  a  new  one  with  little 
expense. 

124 


It  is  used  in  many  of  the  best  types  of  stationary  high-speed  engines, 
because  it  is  simple,  light  and  perfectly  balanced.  It  is  also  much  used 
in  marine  engines,  especially  for  the  high-pressure  cylinders. 


Q.  54.  (1900-01.)     What  is  meant  by  a  poppet  valve? 

How  is  the  lift  determined? 

What  is  meant  by  a  variable  cut-off  gear? 

Explain  the  Stephenson  link  motion  by  the  means  of  a  sketch  fur- 
nished. 

Ans.  54.  Poppet  valves  are  those  that  open  by  rising  from  their 
seats.  They  are  extensively  used  for  the  water  distribution  in  pumps, 
and  also  in  gas  engines.  In  steam  engines  their  use  has  been  restricted 
to  slow-running  marine  engines. 

The  lift  of  such  valves,  if  single,  would  be  about  one-quarter  of  their 
diameter;  if  double,  about  one-eighth;  in  either  case  would  give  an 
area  equal  to  the  steam  port. 

Reversing  the  direction  of  motion  of  an  engine  is  accomplished  by 
a  device,  invented  by  Stephenson,  by  means  of  which  the  engine  may  be 


THE   STEPHENSON  LINK. 

not  only  reversed,  but  the  cut-off  raised  to  any  desired  extent  in  a  very 
simple  manner. 

This  is  called  the  link  motion. 

It  consists  of  two  eccentrics,  with  straps  and  rods.  The  eccentrics 
are  so  placed  that  when  one  is  in  the  right  position  for  the  engine  to 
move  forward,  the  other  is  in  the  position  to  run  backwards.  By  rais- 
ing or  lowering  the  link,  motion  will  b^  communicated  to  the  valve  and 
the  engine  will  move  forward  or  backward  as  the  case  may  be. 

The  result  of  this  combination  is  that  the  link  receives  a  reciprocat- 
ing motion  in  its  center,  since  when  one  eccentric  is  moving  the  end  of 
the  link  in  one  direction  the  other  eccentric  is  moving  the  other  end 
of  the  link  in  the  other  direction;  so  that  the  link  will  have  nearly 
the  same  motion  communicated  to  it  as  if  it  were  suspended  from  a 
pivot  at  its  center. 

The  horizontal  motion  communicated  to  the  link  by  the  joint  action 
of  the  eccentrics  is  a  minimum  at  the  center  of  its  length,  which  is 
equal  to  twice  the  linear  advance  and  it  increases  towards  its  extremi- 
ties, being  nearly  equal  at  either  extremity  to  the  motion  which  would 
be  imparted  to  it  by  the  eccentric  at  that  extremity  alone  without  re- 
gard to  the  other. 


The  valve  rod  is  attached  to  a  block  which  slides  in  the  link,  and 
the  position  of  this  block  is  varied  by  the  means  of  a  combination  of 
rods  and  levers,  attached  in  some  cases  to  the  block  and  in  others  to 
the  link  itself.  In  either  case  the  link  is  suspended  by  a  rod  at  some 
point,  and  the  length  of  the  rod,  as  well  as  the  location  of  the  point 
on  the  link  to  which  it  is  attached,  have  an  important  influence  on  the 
motion  of  the  link. 

The  travel  of  the  valve  depends  upon  the  distance  of  the  block  from 
the  center  of  the  link.  By  moving  the  link  up  or  down  on  the  block 
or  the  block  up  or  down  in  the  link,  the  travel  of  the  valve  may  be 
increased  or  diminished.  The  central  position  corresponds  to  no  mo- 
tion whatever,  while  the  nearer  the  block  is  to  either  eccentric  the 
more  its  motion  will  be  under  the  control  of  that  eccentric.  Now, 
since  the  travel  of  the  valve,  other  things  being  equal,  determines  the 
point  of  cut-off,  it  follows  that  the  degree  of  expansion  raises  with  the 
position  of  the  block  relative  to  the  link. 

In  the  sketch,  for  example,  suppose  the  front  eccentric  is  set  for 
forward  motion  and  the  back  eccentric  for  reverse  motion.  The  link 
is  suspended  by  a  rod  (not  shown),  and  in  the  position  represented  in 
the  sketch  the  block  is  at  the  top,  and  it  is  therefore  entirely  under 
control  of  the  forward  eccentric.  Consequently  the  engine  is  going 
forward  and  the  steam  is  cut  off  at  the  latest  possible  moment,  the 
exact  point  of  the  stroke  when  the  cut-off  takes  place  depending  on 
the  lap  and  other  dimensions  of  the  valve  gear.  Now,  as  the  link  is 
raised  the  travel  of  the  valve  decreases  and  cut-off  takes  place  earlier, 
until  the  block  is  in  the  center,  when  there  is  not  enough  travel  to  un- 
cover the  ports,  and  the  engine  comes  to  rest.  It  is  obvious  without 
further  explanation  that  if  the  block  is  placed  in  the  several  positions 
controlled  by  the  back  eccentric  that  the  effect  will  be  the  same,  only 
in  the  reverse  or  opposite  direction. 

The  term  "full-gear  forward"  means  that  the  link  is  dropped  to  its 
full  extent;  while  "full-gear  backward"  means  that  the  link  is  lifted  to 
its  full  extent 

When  the  link-block  stands  directly  under  the  saddle-plate  both  parts 
are  closed,  and  neither  admission  nor  exhaust  can  take  place.  The 
distance  between  the  block  and  the  end  of  the  link  when  in  full  gear 
is  termed  the  clearance. 

The  radius  of  the  link  is  the  distance  from  the  center  of  the  driving- 
axle  or  the  shaft  upon  which  the  eccentric  is  located,  to  the  center  of 
the  link,  while  the  link  itself  is  segment  of  the  circle  of  that  diameter. 
The  length  may  be  greater  or  less,  but  any  variation  from  these  pro- 
portions will  give  more  lead  at  one  end  than  at  the  other  while  working 
steam  expansively,  but  the  radius  may  be  several  inches  shorter  or 
longer  without  materially  affecting  the  motion. 

The  vital  point  in  designing  a  valve-link  motion  is  the  point  of 
suspension  of  the  link.  If  suspended  from  the  center  it  will  invaria- 
bly cut  off  steam  sooner  in  the  forward  stroke  than  in  the  backward 
stroke,  while  working  expansively.  It  is  customary  to  suspend  the  link 
at  a  point  which  is  most  used  in  the  running  of  the  engine. 

The  ease  and  facility  with  which  the  link  may  be  handled  is  a  very 
important  feature  in  its  favor. 

Q.  55.  (1900-01.)     What  is  the  function  of  a  governor? 

Explain  the  action  of  a  single  throttling  governor.  What  are  its 
principal  defects? 

In  high  speed  machine  why  is  the  shaft  governor  used  so  exten- 
sively? 

Explain  the  action  of  a  single  form  of  a  fly-wheel  governor? 

Ans.  55.    The  function  of  a  governor  is  to  regulate  or  govern  the 

speed  of  the  engine,  admitting  the  requisite  amount  of  steam  to  do  the 

work,  sustaining  the  speed  uniform  or  within  a  small  percentage  of  it. 

s  subject  of  regulating  the  speed  of  steam  engines  has  of  late  years 

126 


received  no  little  attention  from  engineers  and  practical  inventors,  and 
as  a  result  various  kinds  of  governors  have  been  introduced. 

Governors  when  attached  to  throttle-valves,  work  under  circum- 
stances that  necessitate  the  use  of  openings  for  the  passage  of  the 
steam  that  are  too  small  in  area,  so  much  so  that  the  useful  effects  of 
the  steam  are  considerably  diminished.  On  this  depends  the  ill  repute 
of  throttling  engines  as  compared  with  those  which  regulate  with  gov- 
ernor-controlled valve  motions  or  variable  cut-off. 

If  the  valve  of  a  governor  has  too  large  openings  it  will,  owing  to 
the  unsteady  action  of  the  governor,  admit  too  large  a  quantity  of  steam 
and  cause  a  jumping  of  the  engine;  then  in  trying  to  shut  off  the  extra 
amount  it  shuts  it  all  off;  in  fact,  the  governor  cannot  fix  it  exactly 
right,  being  incapable  of  delicate  changes. 

This  difficulty  is  best  met  by  making  the  openings  in  the  valve  of 
peculiar  shape,  so  that  they  open  and  close  in  a  ratio  different  from  that 
of  the  governor. 

The  principle  of  centrifugal  force,  as  embodied  in  the  old  fly-ball 
governor  of  Watt,  has  been  more  resorted  to  than  any  other;  but,  aside 
from  this,  the  governor  has  been  so  improved,  altered,  and  reconstruct- 
ed since  his  time  as  to  be  almost  unrecognizable;  but  still  the  old  prin- 
ciple remains,  and  also  the  prominent  defects  which  so  materially  inter- 
fere with  its  efficiency. 

The  principal  defects  are  three  (3)  in  number.  First  is  the  friction 
which  arises  from  the  joints;  second,  defect  is  due  to  the  fact  that  the 
balls  as  they  assume  different  positions  in  keeping  with  the  speed  with 
which  they  revolve  are  obliged  to  rise  and  fall.  This  is  necessary  in 
order  that  the  resistance  which  the  weights  offer  to  centrifugal  force 
should  constantly  increase.  If  it  did  not  so  increase,  the  weights,  when 
once  started  from  their  position  of  rest,  would  instantly  go  to  the 
extreme  limit  of  motion.  The  rising  of  the  balls  shortens  the  distance 
which  they  are  allowed  to  move  for  a  given  variation  by  bringing  the 
centers  of  the  balls  and  arms  on  which  they  swing  into  a  straight  line, 
so  that  a  variation  which  moves  the  balls  a  given  distance  upward,  if 
it  occurs  again,  will  not  move  them  nearly  so  far  in  the  same  direc- 
tion. Again,  the  same  force  that  would  support  the  balls  in  any  plane 
would  not  raise  them  to  that  plane  from  a  lower  one.  So  between  fric- 
tion, which  destroys  the  delicate  power  that  the  balls  assume  under  a 
slight  change,  and  the  necessity  for  a  large  change  to  overcome  their 
inertia,  it  is  almost  impossible  to  obtain  a  degree  of  regulation  which 
would  be  equal  to  all  requirements. 

The  third  defect  in  governors  on  throttling  engines  is  that  the 
valve  stem  has  of  necessity  to  pass  through  steam-tight  packing  boxes. 
There  is  also  friction  on  the  governor  valve  necessary  to  overcome  the 
power  required  to  move  the  valve-stem  through  all  its  bearings,  guides, 
stuffing-boxes,  etc.,  under  the  pressure  of  steam. 

Were  it  possible  to  construct  a  governor  for  throttling  engines  which 
would  approach  in  practice  what  theory  would  demonstrate  the  fly-ball 
or  centrifugal  governor  would  be  a  perfect  regulator;  but  this  appears 
under  mechanical  laws  impossible. 

By  the  use  of  insochronous  governors,  which  would  not  admit  of  any 
variation  of  speed,  but  would  be  in  equilibrium  at  any  speed,  whether 
the  balls  were  up  or  down  or  in  any  other  position,  the  defects  of  the 
common  governor  were  supposed  to  be  obviated;  but  it  was  found  by 
experience  that  power  and  stability  were  necessary,  and  isochronism 
in  its  strict  sense  unattainable. 

In  the  fly-wheel  governor  these  defects  have  been  partially  elimin- 
ated, and  it  has  been  found  that  much  closer  regulation  is  attainable 
with  the  aid  of  governors  of  this  class  and  also  that  there  is  less 
variation  of  speed  during  each  single  stroke. 

For  these  reasons  nearly  all  modern  steam  engines,  excepting  the 
roughest  type,  are  equipped  with  governors  which  act  upon  the  cut-off. 
Jt  will  be  readily  understood  that  the  tension  of  the  springs  and  the 

127 


weights  may  be  so  adjusted  as  to  maintain  a  position  of  equilibrium 
and  thus  produce  an  approximately  uniform  speed  of  rotation.  With 
the  aid  of  a  well-designed  governor  of  this  type  it  is  possible  to  regu- 
late the  engine  to  within  1  per  cent  of  its  rated  speed  from  full  load  to 
no  load  or  no  load  to  full  load. 

The  action  of  the  fly-wheel  governor,  which  is  the  type  used  on  most 
high-speed  engines,  may  be  described  as  follows: 

The  arms  carrying  weights  are  pivoted  to  the  fly-wheel  and  springs 
are  attached  to  these  arms  and  the  rim  of  the  wheel,  the  links  con- 
necting the  arms  and  collar,  which  is  placed  loosely  on  the  main  shaft, 
which  carries  the  eccentric.  When  the  fly-wheel  revolves  there  is  a 
tendency  for  the  weights  to  travel  outward — that  is,  away  from  the 
shaft — and  the  centrifugal  force  which  actuates  to  move  in  that  direc- 
tion is  counteracted  by  the  tension  of  the  springs.  If  the  speed  in- 
creases this  tension  is  partially  overcome  and  the  weights  move  out- 
ward, and  in  so  doing  shift  the  eccentric  and  changing  its  eccentricity, 
angle  of  advance,  or  both,  and  thus  produce  an  earlier  cut-off  in  the 
cylinder,  or,  as  the  case  may  be,  this  operation  may  be  reversed  and  a 
later  cut-off  take  place. 

Governors  should  be  kept  perfectly  clean  and  free  from  accumula- 
tions induced  by  the  use  of  inferior  oils,  as  such  gummy  substances 
have  a  tendency  to  interfere  with  the  easy  movement  of  the  different 
parts;  parts  working  through  packings  should  be  frequently  packed, 
that  the  packing  may  be  kept  soft. 


Q.  56.  (1900-01.)  What  is  the  effect  on  the  steam  distribution  if  the 
cut-off  is  varied  by  altering  the  angular  advance  only  or  the  eccentricity 
only? 

In  what  form  of  valve  gear  is  it  not  necessary  to  vary  both  the 
angular  advance  and  the  throw  of  the  eccentric? 

Ans.  56.  In  the  case  of  a  single  valve  operated  by  a  shaft  governor, 
if  the  angle  of  advance  only  is  altered,  the  cut-off  may  be  varied  to  suit 
the  conditions  of  load,  but  in  that  case  the  lead  will  also  vary  in  such  a 
way  that  it  would  increase  as  the  cut-off  decreased.  If,  on  the  other 
hand,  the  regulation  is  performed  by  varying  the  eccentricity  alone,  the 
reverse  would  take  place,  and  hence  for  shaft  regulation  in  connection 
with  a  single  valve  it  becomes  necessary  to  vary  the  eccentricity  and 
angle  of  advance  simultaneously. 

This  is  not  the  case  where  separate  distribution  and  expansion  valves 
are  used,  as  in  the  "Buckeye  Engine,"  because  the  admission  and  ex- 
haust closure  are  affected  by  the  distribution  valve,  and  the  governor 
acts  only  on  the  cut-off. 


Q.  57.  (1900-01.)  How  would  you  calculate  the  diameter  of  gover- 
nor pulleys? 

What  is  the  maximum  variation  in  speed  allowable  with  a  good  shaft 
governor? 

Explain  the  action  of  the  governor  in  the  Corliss  type  of  engines. 

Why  can  Corliss  engines  not  operate  at  a  high  rotative  speed? 

Ans.  57.    To  find  the  diameter  of  the  governor  shaft-pulley— 

Multiply  the  number  of  the  revolutions  of  the  engine  by  the  diameter 
of  the  engine  shaft  pulley  and  divide  the  product  by  the  number  of  revo- 
lutions of  the  governor. 

To  find  the  diameter  of  the  engine  shaft  pulley — 

Multiply  the  number  of  revolutions  of  the  governor  by  the  diameter 
of  the  governor  shaft  pulley  and  divide  the  product  by  the  number  of 
revolutions  of  the  engine. 

128 


The  maximum  variation  in  speed  in  the  modern  first-class  high-speed 
engines  should  not  exceed  more  than  2  per  cent.  Many  engines  are 
guaranteed  that  from  full  load  to  no  load  the  variation  will  not  exceed 
1  per  cent. 

In  Corliss  engines  the  economy  in  the  use  of  steam  and  the  close 
regulation  which  has  been  attained  certainly  makes  this  class  of  engines 
very  desirable  for  a  great  many  purposes. 

Owing  to  the  method  of  effecting  the  cut-off  they  are,  however,  lim- 
ited as  to  speed  of  rotation,  and  it  is  consequently  customary  to  secure 
the  advantages  which  are  to  be  derived  from  high-piston  speeds  by 
making  the  stroke  very  long.  Of  course,  this  renders  them  unfit  for 
direct  coupling  to  electrical  and  other  machinery  which  needs  to  run 
at  a  comparatively  great  number  of  revolutions  per  minute,  but  there 
are  many  other  uses  to  which  the  Corliss  can  be  put  to  advantage. 


Q.  58.  (1900-01.)  What  are  the  characteristics  of  high  speed  auto- 
matic cut-off  engines? 

For  what  class  of  service  are  they  especially  adapted?    Why? 

What  is  the  steam  consumption  per  horse-power  in  this  class  of  en- 
gine? 

Ans.  58.  The  class  of  engines  known  as  the  high-speed  automatic 
cut-off  type,  which  now  comprise  a  large  variety  of  designs,  owes  its 
development  largely  to  the  unusual  growth  of  electric  lighting  and 
power,  isolated  plants  and  electric  railway  service.  Engines  of  this 
class,  while  applicable  under  various  conditions,  are  designed  primarily 
to  meet  the  requirements  which  the  nature  of  this  service  imposes. 

An  engine  used  for  driving  a  lighting  or  power  generator  is  con- 
stantly subjected  to  sudden,  and  very  often  considerable,  variations  of 
load,  and  must  under  these  circumstances  maintain  a  constant  or  nearly 
constant  speed.  It  must  also  be  economical  in  the  use  of  steam,  run 
at  a  comparatively  high  rotary  speed,  and  be  simple  in  design.  In 
general,  these,  briefly,  are  the  conditions  which  have  evolved  the  high- 
speed automatic  cut-off  engine,  these  engines,  while  not  so  economical 
in  the  use  of  steam  as  the  Corliss  type,  are  vastly  better  than  the  old 
throttling  engines.  They  consume  from  thirty  to  thirty-five  pounds  of 
steam  per  horse-power  hour,  when  operating  under  an  initial  pressure 
of  eighty  pounds  and  cutting  off  at  one-quarter  stroke,  while  where 
compounded,  which  is  frequently  done,  their  steam  consumption  is  about 
twenty-five  pounds. 

Q.  59.  (1900-01.)  Explain  what  is  meant  by  a  "four- valve"  engine, 
and  why  it  is  more  economical  than  a  single-valve  engine? 

What  are  the  advantages  and  disadvantages  of  single-acting  engines? 

Ans.  59.  The  four-valve  engine  is  a  type  which  has  been  designed  to 
combine  some  of  the  principal  advantages  of  the  Corliss  type  and  the 
high-speed  engines.  It  resembles  the  high-speed  type  in  that  the  cut- 
off is  varied  to  meet  the  changes  in  load  by  means  of  a  shaft  governor, 
and  it  resembles  the  Corliss  engine  in  having  separate  passages  for  the 
admission  and  the  exhaust,  thus  avoiding  the  losses  inherent  in  single- 
valve  engines  and  at  the  same  time  retaining  the  advantages  to  be 
derived  from  high  rotative  speed,  which  may  be  had  on  account  of  the 
absence  of  the  releasing  mechanism. 

There  being  separate  parts  -for  admission  and  exhaust,  the  fresh 
steam  does  not  come  in  contact  with  the  surfaces  which  have  been 
previously  cooled  by  the  exhaust,  hence  the  condensation  of  the  steam 
i§  reduced  to  the  minimum. 

Some  of  the  advantages  of  the  single-acting  engine  are  high  rotative 
speed,  simplicity  of  design  and  the  economy  that  has  been  obtained  in 
steam  consumption. 

The  results  which  are  attained  in  this  respect  are  due  to  high  speed, 
multiple  expansion,  quick  cut-off  and  short  steam  passages.  The  lim- 

129 


ited  amount  of  floor  space  occupied  makes  its  application  very  desira- 
ble where  the  available  space  is  limited.  The  two  principal  designs  of 
this  class  of  engines  are  the  "Willans"  and  the  Westinghouse. 

These  engines  are  well  adapted  for  a  variety  of  uses.  They  are 
especially  valuable  in  location  where  the  engine  is  exposed  to  dust, 
since  the  working  parts  are  almost  completely  enclosed  in  the  casing. 
The  high  speed  and  close  regulation  make  it  useful  also  for  electric 
lighting  and  railway  service,  although  it  is  not  nearly  so  economical 
in  the  use  of  steam  as  many  other  types  of  engines. 


Q.  60.  (1900-01.)  What  are  the  most  frequent  causes  of  knocking 
in  steam  engines? 

How  may  knocks  be  located? 

What  are  the  remedies  for  different  kinds  of  knocks? 

Explain  how  an  engine  with  a  fixed  eccentric  may  be  reversed. 

Ans.  60.  The  most  frequent  cause  of  knocking  in  steam  engines  are 
lost  motion  in  the  cross-head,  wrist  and  crank  pin  boxes,  looseness  in 
the  pillow-block  or  main  bearing  boxes,  looseness  of  the  piston  rod  or 
follower-plate,  the  crank  pin  or  crank  shaft  being  out  of  line  with  the 
cylinder  or  the  wrist  pin,  crank  pin  or  main  bearing  journal  being  worn 
oval;  the  slide  valve  having  too  much  lead  or  not  enough;  the  exhaust 
opening  being  too  soon  or  too  late;  the  valve  being  badly  proportioned 
or  the  exhaust  passage  outside  of  the  cylinder  being  contracted. 

Other  causes  are  shoulders  being  worn  in  each  end  of  the  cylinder, 
in  consequence  of  packing  rings  not  traveling  over  the  counter-bore  at 
each  end  of  the  stroke;  or  shoulders  being  worn  on  the  guides,  result- 
ing from  cross-head  shoes  not  overlapping  them  when  the  crank  is  on 
the  dead  center;  the  piston  not  having  sufficient  clearance  at  either 
end  of  the  cylinder,  in  consequence  of  its  being  altered  by  taking  up 
the  lost  motion  in  the  boxes;  there  not  being  sufficient  draught,  in  the 
keys  to  take  up  the  lost  motion  in  connecting-rod  boxes;  the  packing 
being  screwed  too  tight  round  the  piston-rod;  excessive  cushioning, 
resulting  from  the  leaky  condition  of  the  piston,  which  allows  the  steam 
to  occupy  the  space  between  the  cylinder  and  piston-head,  as  the  crank 
approaches  the  center,  thereby  subjecting  the  engine  to  an  enormous 
strain,  as  at  this  part  of  the  stroke  the  fly-wheel  is  traveling  very  fast 
and  the  crank  moving  very  slowly;  lost  motion  in  the  connection  by 
which  the  slide-valve  is  attached  to  the  rod. 

Engines  out  of  line  frequently  knock  sideways  at  the  half-stroke, 
but  most  generally  at  the  outward  and  inward  upper  or  lower  dead 
center,  as  these  are  the  points  that  the  greatest  strain  is  thrown  on 
the  bearings  in  consequence  of  the  direction  of  the  connecting-rod  hav- 
ing to  be  reversed.  The  foregoing  causes  of  knocking  in  engines  con- 
stitute the  principal  ones. 

Knocks  arising  from  lost  motion  in  any  of  the  revolving,  recipro- 
cating or  vibrating  parts  of  an  engine  may  be  located  by  placing  the 
fingers  on  the  part,  while  the  cross-head  is  being  moved  back  and 
forth  on  the  guides  by  the  starting  bar;  but  knocks  induced  by  the  valve 
opeing  or  closing  too  soon,  by  the  contraction  of  the  exhaust,  or  by  the 
valves  being  improperly  set,  are  the  most  difficult  to  discover,  as  they 
are  different  from  those  induced  by  lost  motion,  the  sound  being  a  dull, 
heavy  thud  in  many  instances,  causing  the  engine  building  and  even 
the  foundation  to  vibrate  at  each  stroke. 

While  an  intelligent  and  careful  search  will  in  most  cases  result  in 
successfully  locating  the  knock,  some  will  for  a  time  baffle  the  most 
expert  engineer.  There  are  instances  where  the  indicator  has  been 
applied  in  order  to  determine  the  precise  location  of  the  knock  or 
thud. 

130 


Remedies  for  knocking  in  steam  engines:  While  it  is  possible  in 
most  cases  to  locate  the  knocking,  it  is  hardly  possible  to  prescribe  a 
remedy  for  all,  as  in  many  instances  it  must  arise  out  of  and  be  deter- 
mined by  circumstances  of  the  individual  case. 

The  most  practical  method  of  remedying  the  knocking  induced  by 
the  crank-pin  being  out  of  line,  is  to  place  the  crank-shaft  at  right  an- 
gles with  the  center  of  the  cylinder,  remove  the  old  crank-pin,  rebore 
the  hole  so  as  to  bring  the  center  of  the  new  pin  perfectly  in  line  with 
the  axis  of  the  cylinder,  and  replace  the  old  pin  with  a  new  one. 

The  knocking  induced  by  the  wrist-pin  and  crank-pin  becoming  worn 
oval,  may  be  remedied  by  filing  them  perfectly  round;  but  the  knocking 
caused  by  the  crank-shaft  journal  being  worn  out  of  round  is  very  dif- 
ficult to  remedy;  in  fact,  there  is  no  remedy  for  it  except  remove  the 
shaft,  true  it  up  in  a  lathe  and  refit  the  boxes;  this  operation  is  attend- 
ed with  considerable  difficulty,  more  especially  when  the  engine  is  large. 

Knocking  in  boxes  on  the  crank-pin  and  cross-head  or  valve  rod  may 
be  remedied  by  refitting  the  boxes,  readjusting  the  keys  or  by  putting 
a  liner  behind  or  in  front  of  the  boxes  when  there  is  not  sufficient 
draught  in  the  keys  and  gibs. 

Knocking  in  the  steam-chest  caused  by  looseness  in  valve  connec- 
tions may  be  remedied  by  readjusting  the  jam-nuts  or  the  yoke.  Knock- 
ing arising  from  this  cause  manifests  itself  more  frequently  when  the 
steam  is  shut  off  from  the  cylinder,  preparatory  to  stopping  the  engine, 
than  when  the  engine  is  running;  the  lost  motion  is  taken  up  in  the 
valve  connections  by  the  pressure  of  the  steam  on  the  back  of  the 
valve. 

Knocking  in  the  piston  is  generally  caused  by  the  rod  becoming 
loose  in  the  head,  and  if  it  continues  for  any  length  of  time  it  destroys 
the  fit  of  the  rod  in  the  hole. 

The  only  practical  remedy  is  to  remove  the  rod,  rebore  the  hole, 
bush  it  or  thicken  the  rod  at  that  point  by  welding,  and  fit  to  the  head 
after  the  hole  is  rebored  perfectly  true. 

Knocking  in  follower-plate  is  generally  caused  by  bolts  being  too 
long  or  from  dust  allowed  to  accumulate  in  the  holes,  which  prevents 
them  from  entering  sufficiently  far  to  take  up  the  lost  motion  in  the 
plate.  Remedied  by  cleaning  out  the  holes  or  shortening  the  bolts. 

Knocking  caused  by  shoulders  becoming  worn  in  cylinders  at  each 
end  can  be  remedied  by  reboring  the  cylinder,  making  the  counterbore 
sufficiently  deep  that  a  part  of  one  of  the  rings  will  overlap  it  at  each 
end  of  the  stroke. 

Shoulders  worn  on  guides  can  be  remedied  by  planing  the  guides  and 
making  the  shoes  sufficiently  long  that  they  will  overrun  the  guides 
when  the  crank  is  at  either  center. 

The  knocking  induced  by  any  of  the  foregoing  causes  is  generally  a 
source  of  great  annoyance,  as  any  attempt  to  adjust  the  boxes  on  cross- 
head  or  crank-pin  or  the  piston-packing  in  cylinder  generally  aggravates 
the  cause  of  the  knocking,  as  any  adjustment  of  the  connecting  rod 
boxes  alters  the  position  of  the  piston  in  cylinder  and  the  cross-head 
on  guides,  and  causes  them  to  strike  harder  against  the  shoulders. 

Knocking  caused  by  the  valve  or  valves  being  improperly  set  may 
be  remedied  by  removing  the  bonnet  of  steam-chest  and  adjusting  the 
valve,  so  that  it  may  move  uniformly  on  its  seat,  thereby  giving  the 
valve  the  same  and  proper  amount  of  lead  at  each  end  of  the  stroke; 
then  if  the  valve  is  well  proportioned  and  the  connections  thoroughly 
fitted  and  skillfully  adjusted  there  is  no  reason  why  the  engine  should 
knock  from  this  cause. 

The  knocks  arising  from  bad  proportion  in  the  valves  and  steam 
passages  are  the  most  difficult  of  all  to  remedy,  as  they  are  inherent  in 
the  machine. 

Hov/  to  reverse  an  engine: 

Place  the  crank  orr  the  dead  center  and  remove  the  bonnet  from  the 


steam  chest;  observe  the  amount  of  lead  or  opening  that  the  valve  has 
on  the  steam  end;  then  loosen  the  eccentric  and  turn  it  around  on  the 
main  shaft  in  the  direction  in  which  it  is  intended  the  engine  should 
run  until  the  valve  has  the  same  amount  of  lead  on  the  other  end. 
The  engine  should  then  be  turned  on  the  other  center  for  the  purpose 
of  equalizing  the  lead;  the  crank  should  also  be  placed  half-stroke,  top 
and  bottom,  for  the  purpose  of  determining  whether  the  part  opening 
Is  the  same  in  both  positions.When  the  crank  is  half-stroke  the  center 
of  crank-pin  is  plump  with  the  center  of  crank-shaft. 


132 


Pumps,   Compressors,   Hydraulics,   Refrig- 
eration, Heaters,  Condensers, 
Injectors,  Etc. 


Q.  8.   (1896-7.)     How  to  properly  set  duplex  steam  pump  valves? 

Ans.  8.  Place  steam  pistons  in  the  center  of  their  travel,  which 
will  bring  the  rocker  arms  perpendicular  to  rods.  Place  slide  valve 
over  center  of  steam  ports.  The  lost  motion  in  pumps  of  small  size 
that  usually  have  non-adjustable  blocks,  on  valve  stems,  should  meas- 
ure the  same  on  each  side  of  blocks,  between  valve  lugs.  If  lost  mo- 
tion is  equal  the  valves  are  set.  Move  one  valve  so  as  to  admit  steam 
before  putting  on  steam  chest  covers.  On  the  medium  and  larger  sizes, 
where  adjustable  locknuts  are  used  on  valve  stems,  there  are  no  fixed 
rules  for  the  exact  amount  of  lost  motion,  which  governs  the  length  of 
stroke,  and  this  can  only  be  determined  satisfactorily  by  trial  and 
careful  adjustment.  In  the  large  pumps  with  the  outside  slotted  link 
and  sliding  block,  where  the  valve  movement  is  coincident  in  time  and 
amount  to  valve  rod  movement  the  distance  from  each  end  of  link 
block  to  each  end  of  link  slot  should  measure  the  same  when  slide 
valves  are  central  over  ports. 


Q.  9.  (1896-7.)  HP.  required  to  raise  300  tons  water  150  ft.  in  one 
and  one-half  hours,  barring  friction? 

Ans.  9.  300  (tons)  X  2,000  (Ibs.)  X  150  (ft.)  -r-  1.5  (hours)  -=-  60 
(min.)  -T-  33,000  (ft.  Ibs.)  =  30  10/33  HP. 


Q.  12.  (1896-7.)  How  to  find  steam  cylinder  diameter  for  direct 
acting  boiler  feed  pump? 

Ans.  12.  Find  the  maximum  amount  of  water  required.  Find  the 
size  of  the  plunger  that  will  displace  this  amount  of  water  at  a  fair 
piston  speed  and  make  the  steam  end  2*4  times  the  area  of  the  water 
end.  [Pump  cylinder  should  allow  2%  to  15%  for  slip.] 


Q.  13.  (1896-7.)  How  many  tons  of  ice  (spec.  gr.  .9188)  to  fill  a 
room  15  ft.  X  16  ft.  X  10  ft.? 

Ans.  13.  15  (ft.)  X  16  (ft.)  X  10  (ft.)  X  62.5  (Ibs.)  X  .9188  (spec, 
grav.)  -r-  2,000  (Ibs.),  =  68.91  (tons).  [Ice  should  be  packed  with  1" 
strips  of  wood  between,  hence  allowance  must  be  made  for  the  strips.] 


Q.  16.  (1896-7.)  What  is  efficiency  of  direct  acting  steam  pump 
from  coal  in  furnace? 

Ans.  16.  A  direct  acting  steam  pump  requires  about  120  Ibs.  steam 
per  hour  per  HP.  Assuming  that  the  boiler  evaporates  10  Ibs.  of  watef 
per  pound  of  coal  there  would  be  12  Ibs:.  coal  used  per  HP.  per  hour. 

133 


The  number  of  foot  pounds  used  per  hour  in  the  furnace  would  he 
13  000  X  778  X  12,  assuming  the  value  of  each  pound  of  coal  to  be  13,000 
heat  units.  This  divided  by  60  would  give  the  heat  units  used  per  min- 
ute and  this  result  multiplied  by  100  and  divided  into  33,000  would 
give  the  percentage  of  efficiency. 

=  1.66  %  efficiency. 


Q.  30.  (1896-7.)  What  is  the  cause  of  water  hammer  in  the  dis- 
charge pipes  of  pumps,  and  how  is  it  avoided? 

Ans.  30.  The  inertia  of  the  water,  which,  when  in  motion,  tends 
to  continue  in  motion,  after  the  impelling  force  has  ceased  to  act  and 
its  energy  of  motion  is  expended  in  giving  a  blow  to  the  containing 
p:pe.  Any  means  which  will  induce  an  uninterrupted  flow,  or  which 
will  bring  the  water  gradually  to  rest,  will  prevent  water  hammer. 

The  first  result  can  be  obtained  by  the  use  of  a  large  air  chamber, 
or,  in  the  case  of  a  duplex  pump,  by  adjusting  the  steam  valves  so  that 
one  piston  begins  its  stroke  before  the  other  has  stopped. 

The  second  result  (gradually  checking  the  flow)  can  be  secured  by 
placing  an  air  chamber  on  the  discharge  pipe  near  the  end  at  the  pump 
or  by  having  the  pump  cushion  in  such  a  manner  as  to  bring  the  piston 
gradually  to  rest  and  not  keep  up  its  speed  to  the  extreme  end  of  the 
stroke  and  then  abruptly  stop. 

The  admission  of  air  with  the  water  entering  will,  in  some  cases, 
prevent  the  trouble  and  a  small  snifting  valve,  opening  inward,  placed 
on  the  suction  pipe  near  the  pump,  will  in  case  the  discharge  pressure 
Is  low,  sometimes  stop  the  hammer. 


Q.  48.  (1896-7.)  What  is  the  most  economical  for  feeding  boilers,  a 
direct  acting  pump,  a  power  pump,  or  an  injector?  Has  one  any  ad- 
vantage over  the  others  under  certain  conditions?  If  so,  when  and 
how? 

Ans.  48.  In  cases  where  the  exhaust  from  engines  can  be  utilized 
for  heating  the  feed  water  the  injector  cannot  be  used  with  economy, 
because,  first,  if  an  open  heater  is  used  the  injector  will  not  take  the 
water;  second,  if  a  closed  heater  is  used  the  feed  will  not  take  up 
much  of  the  heat  in  the  exhaust  steam  for  the  reason  that  the  feed 
water,  after  passing  through  the  injector,  is  nearly  as  hot  as  the  ex- 
haust steam  itself. 

A  power  pump  driven  by  a  belt  from  the  shafting  is  more  economical 
as  a  boiler  feeder  than  either  a  direct  acting  pump  or  an  injector.  It 
has,  however,  this  disadvantage,  that  the  shaft  must  run  in  order  to 
get  water  into  the  boiler.  On  account  of  this  it  is  not  generally  used. 

The  direct  acting  steam  pump,  while  it  is  not  as  economical  as  the 
power  pump,  is  independent  of  the  engine,  and  can  be  placed  in  any 
part  of  the  boiler  or  engine  room,  and  is  ready  to  start  as  soon  as 
there  are  a  few  pounds  pressure  of  steam  in  the  boiler. 

The  injector,  while  it  is  still  less  economical  than  either  the  power 
pump  or  direct  acting  pump,  is,  in  cases  where  the  feed  water  is  cold 
and  no  waste  heat  is  available  for  heating  it,  preferable  and  more  eco- 
nomical than  either  the  steam  pump  or  power  pump,  first,  because  it 
furnishes  hot  feed  water  for  the  boiler,  and,  second,  because  the 
injector,  taken  as  a  pump  and  feed  water  heater,  has  a  very  high  effi- 
ciency. But,  all  things  taken  into  account,  we  consider  the  direct  acting 
steam  pump  the  most  economical  method  of  feeding  boilers. 

Q.  54.  (1896-7.)  How  do  we  find  the  amount  of  injection  water 
required  per  HP  per  hour? 


Ans.  54.  The  amount  of  injection  water  required  per  indicated 
horse-power  will  depend  upon,  first,  the  temperature  of  the  injections; 
second,  the  economy  of  the  engine;  third,  the  final  temperature  of  the 
hot  well.  The  unit  of  water  required  per  unit  of  steam  condensed  can 
be  found  by  the  following  formula: 

Let  I  =  temperature  of  injection. 

Let  U  —  temperature  of  discharge. 

Let  S  =  total  heat  of  steam  discharged  at  release. 
S— D 

=  lbs.  of  water  per  Ib.  of  steam  condensed. 

D— I 

Example:  Temperature  of  injection,  60°;  temperature  of  discharge, 
120°;  total  heat  above  32°  in  1  Ib.  of  steam  of  20  Ibs.  pressure  equals 
1150  units;  then — 

1150  —  120       1030 

= =  17.16  Ibs.  water  per  pound  of  steam  condensed. 

120  —  60  60 

If  20  Ibs.  of  steam  are  used  per  horse-power  per  hour,  then  20  X  17.16 
=  343.20  Ibs.  of  water  used  per  HP  per  hour. 

This  amount  increases  with  temperature  of  injection  and  inversely 
with  the  temperature  of  the  discharge. 


Q.  81.  (18^6-7.)  A  pumping  engine  works  against  a  gage  pressure 
of  43.4  Ibs.,  and  a  vacuum  of  25"  is  needed  to  raise  water  to  the  pump: 
What  will  be  the  total  height  in  feet  against  which  the  pump  works? 

Ans.  81.     No  temperature  being  given,  we  will  assume  that  1  ft.  of 
water  exerts  a  pressure  of  .434  Ibs.  per  sq.  in.,  also  that  1"  mercury 
equals  1.133  ft.  of  water,  then: 
43.4 

=  100  ft.  on  discharge  side  and  1.133  X  25  inches  vacuum  =  28.325 

.434 

ft.  on  suction  side,  or 
43.4 

h  [1.133  X  25]  =  128.325  ft. 

.434 


Q.  91.  (1896-7.)  An  air  compressor  takes  air  at  atmospheric  pres- 
sure (15  Ibs.)  and  60°  F.,  and  compresses  same  adiabatically  to  120  Ibs. 
gage  pressure.  Find  the  temperature  at  the  end  of  compression.  The 
ratio  of  relative  specific  heats  is  1.4. 

Ans.  91.     T1  =  absol.  temp,  after  compression. 

T  ==  absol.  temp,  before  compression,  or  460°+  60°=  520°. 
P1  =  absol.  pressure  after  compression,  120  +  15  =  135  Ibs. 
P  =  absol.  pressure  before  compression  =  15  Ibs. 

Tl        /P1\    29  /P1\    29  /lSo\    29 

=••     c,T,=T-      =T'=5»-     =520X9." 


/P1\    29  /lSo\    29 

,=T(|)-      =T'=5»Q-     = 


log  9  =  .954243  X  .29  =  .276731  representing  1.8915:   520  X 
1.8915  =  983.32,  total  tempr.,  less  460  =  523.3°. 


Q.  94.  (1896-7.)  Given  a  piston  force  pump  with  a  lever  of  the 
second  order,  having  a  2"  piston  and  a  ram  with  25"  diameter,  pump 
lever  is  5"  from  fulcrum  to  center  of  piston,  and  60"  from  center  of 
piston  to  end  of  lever:  How  much  weight  needed  on  end  of  lever,  to 
equalize  a  load  of  100,000  Ibs.  on  the  ram,  ignoring  friction  or  weight  of 
parts? 

135 


Ans.  94.     252  X  .7854  =  490.8  sq.  in.  area  of  ram. 

22  X  .7854  =  3.1416  sq.  in.  area  of  piston. 
100,000  Ibs. 

=  203.72  Ibs.  pressure  on  each  inch  of  ram  and  piston. 

490.8 

3.1416  X  203.72  =  640.07  Ibs.  total  pressure  on  piston. 
640.07  X  5 

.-. =  49.23  Ibs.,  or  say  50  Ibs. 

65 
Or  a  shorter  method:     252  =  625,  22  =  4. 

100,000  640X5 

=  160.      160X4  =  640.     =  49.23  + 

625  65 


Q.  100.  (1896-7.)  In  a  refrigerating  plant  it  is  required  to  find  the 
thermal  units  of  refrigeration  and  of  condensation,  also  the  efficiency 
of  compression,  the  specific  heat  of  the  liquid  being  1.08,  when  the 
inferior  absolute  pressure  (suction  pressure)  of  the  ammonia  is  40 
Ibs.  to  the  square  inch.  The  temperature  when  it  enters  the  compressor 
is  25°  F.  and  it  is  compressed  to  165  Ibs.  per  square  inch  absolute  with- 
out superheating,  then  condensed  into  a  liquid  of  85°  F.,  then  admitted 
to  the  cooling  coils  evaporated,  and  heated  to  the  initial  state. 

Ans.  100: 

Temperature  absolute  at  40  Ibs r2  =  472 

Temperature  of  liquid  472-460 T2  =    12 

Absol.  tempr.  of  saturation  460+25 : . .  r  =  485 

Superheating r-r2  =   13 

Absol.  tempr.  at  end  of  compression ri  =  545 

Absol.  tempr.  of  saturation  460  +  25 r  =  485 

Temperature  of  liquid  at  PI Tt  =    85 

Fall  of  tempr.  from  Ti  to  T2 T,-T2  =    73 

Heat  absorbed  during  this  fall  1.08  X  73 =   78.84 

Latent  heat  of  evaporation  at  T2 h  =  548 

Condenser  heat  removed   ht  =  502 

Refrigeration  per  Ib.  of  ammonia  548  +  (13  X  0.508) — 78  84.  .h2  =  475  76 
Efficiency  h2 


E=    18.13 


That  is,  for  every  thermal  unit  of  work  done  by  the  compressor 
18.13  thermal  units  will  be  removed  from  the  cold  room.  The  efficiency 
of  a  compressor  is  greatest  when  the  difference  between  the  inferior 
and  superior  temperature  (^  to  r2)  is  the  smallest. 


Q.  38.  (1897-8.)  What  is  the  greatest  height  to  which  a  pump  may 
draw  water  that  is  at  a  temperature  of  191°  F.? 

Ans.  38.  The  pressure  of  the  atmosphere  is  about  14.7  Ibs.  per 
square  inch.  The  pressure  of  steam  at  190°  is  about  9.5  Ibs.  per  square 
inch.  There  could  never  be  a  vacuum  less  than  the  pressure  of  the 
KMT™'  Q  f  °  -fS*,^  Pressure  possibly  available  to  raise  the  water  is 
14.7  —  9.5  =  5.2  Ibs.  per  square  inch.  This  corresponds  to 

5.2 

=  12  ft. 

.434 
nearly.     (See  No.  28  of  last  year's  questions.) 

136 


Q.  39.  (1897-8.)  What  effect  will  the  required  velocity  of  the  water 
in  the  inlet  pipe  to  the  pump  have  on  the  height  to  which  the  water 
may  be  raised,  neglecting  the  friction  of  the  water  on  the  pipe? 

Ans.  39.  It  will  lessen  the  height  by  an  amount  equal  to  the  "head" 
required  to  give  it  the  velocity.  This  head  would  be  about  equal  to 
the  square  of  the  velocity  (in  feet  per  second),  divided  by  64. 

V2 
h  =  —  + 

64. 

A  number  have  answered  No.  39  by  saying  that  the  velocity  of  the 
water  in  the  inlet  pipe  would  have  practically  no  effect  upon  the  height 
to  which  the  water  could  be  drawn.  This  is  no  doubt  so  if  the  velocity 
of  the  water  is  small  and  regular.  Nevertheless  we  believe  that  very 
often  when  a  pump  fails  to  take  water  for  its  entire  stroke  the  reason  is 
to  be  found  in  the  inertia  of  the  water  entering  the  inlet  pipe.  [Air  in 
water  and  the  vapor  of  water  produced  by  the  vacuum,  as  well  as  slip 
of  valves,  all  tend  to  prevent  the  cylinder  from  filling. — Ed.] 


Q.  29.  (1898-9.)  Define  omits  of  ice  making  and  refrigerating  ca- 
pacity. 

Anc.  29.  In  refrigeration,  an  effect  equivalent  to  the  conversion  of 
2,000  pounds  of  water  at  32°  into  a  ton  of  ice  at  32°;  that  is,  the  ab- 
straction of  284,000  B.  T.  U.  is  considered  a  "ton,"  the  latter  word  in 
this  sense  being  a  unit  expressive  of  the  duty  just  described. 


Q.  102.  (1898-9.)  Give  the  piston  speed  you  consider  best  adapted 
for  boiler-feed  pump,  when  same  is  supplying  a  "continuous"  feed. 

Ans.  102.  A  desirable  piston  speed  for  boiler  feed  pump  when  sup- 
plying a  continuous  feed  approximates  50  ft.  per  minute. 


Q.  103.  (1898-9.)  Name  the  type  of  pump  and  dimensions  necessary 
to  supply  the  boilers  referred  to  in  Q.  93. 

Ang.  103.  Dimensions  of  feed  pump  for  boilers  referred  to  in  Ans. 
No.  93  may  be  determined  as  follows: 

Figuring  on  200  HP  as  the  normal  demand  on  boilers  but  providing 
50%  for  overload  =  300  HP. 

Allowing  30  Ibs.  water  per  HP  plus  25%  for  non-efficiency  of  pump 
over  its  nominal  displacement,  gives  40  Ibs. 

300  X  40 

Or =  200  Ibs.  of  feed  water  to  be  "moved"  per  minute. 

60 

Assuming  50  ft.  of  piston  travel  as  per  Ans.  No.  102  we  have  200  -=- 
50  =  4  Ibs.  or  %  gallon  as!  the  required  displacement  per  foot  of  piston 
travel  and  which  is  equivalent  to  3%"  dia.  plunger  for  single  cylinder 
or  2l/2"  dia.  for  double  cylinder  pump.  Stroke  of  pump  may  be  any 
length  common  to  the  trade,  for  the  given  diameters. 

If  conditions  permitted  a  power  pump,  one  of  the  "triple"  type,  would 
be  preferred. 

The  independent  steam  pump  has  an  "emergency  reserve,"  which  is 
equal  to  any  margin  of  speed  that  can  be  properly  attained  above  the  50 
ft.  which  is  usually  figured  on.  When  the  limit  of  speed  in  a  power 
pump  is  fixed  in  its  driving  mechanism,  such  a  reserve  is  not  available 
and  for  that  reason  their  capacity  should  be  somewhat  greater  than 
figured  for  the  steam  pump. 

Power  pumps  may  be  kept  in  constant  motion  and  arranged  to  sup- 
ply either  a  constant  or  variable  feed  by  providing  a  suitable  "by-pass" 
arrangement. 

•'37 


Q.  104.  (1898-9.)  What  would  probably  be  the  size  of  the  pipe  con- 
nections on  any  standard  injector,  capable  of  supplying  boilers  of  200 
HP? 

Ans.  104.  IW  would  be  the  minimum  size  for  the  injector-pipe 
connections  for  the  duty  referred  to  in  this  question. 


Q.  105.  (1898-9.)  Explain  the  general  features  claimed  for  open  and 
closed  feed-water  heaters. 

Ans.  105.  Peed-water  heaters  tend  to  economy  and  obviate  strains 
caused  by  forcing  cold  water  into  hot  boilers.  Ordinarily  the  saving 
of  fuel  amounts  to  about  1%  for  every  11°  of  temperature  added  to  the 
"feed"  by  the  heater. 

Heaters  of  the  "open"  type  are  preferable  for  waters  that  will  pre- 
cipitate solid  matter  at  temperatures  near  the  boiling  point.  Well  con- 
structed heaters  of  this  style  heat  the  feed  water  very  efficiently;  they 
intercept  large  quantities  of  "scale"  making  matter  which  is  an  especial 
advantage  with  certain  waters.  They  can  also  be  thoroughly  inspected 
and  cleaned  when  necessary.  The  exhaust,  mingling  with  the  feed 
water  condenses  a  percentage  of  the  steam  but  also  causes  oils  from 
cylinder  lubrication  to  become  mixed  with  the  feed. 

The  mixing  of  the  oil  and  the  further  disadvantage  of  pumping  hot 
water  do  not  obtain  with  "closed"  heaters,  but  as  both  of  these  so- 
called  "difficulties"  which  attend  open  heaters  are  readily  overcome 
they  are  very  much  in  vogue,  for  the  reasons  which  are  noted. 


Q.  67.  (1899-1900.)  What  is  the  force  against  which  a  pump  works, 
aside  from  boiler  pressure? 

Ans.  67.  Gravity,  or  the  attraction  of  the  earth,  which  prevents 
the  water  from  being  lifted.  This  is  shown  in  the  fact  that  water  can 
be  led,  or  trailed,  an  immense  distance,  limited  only  by  the  friction 
of  the  pump. 

Q.  68.  (1899-1900.)  In  designing  or  purchasing  a  pump,  what  is  a 
safe  rule  as  to  capacity? 

Ans.,  68.  One  should  be  selected  capable  of  delivering  1  cu.  ft.  of 
water  per  H.P.  per  hour;  or,  in  other  words,  about  3  Ibs.  of  water  for 
each  sq.  ft.  of  heating  surface. 


Q.  69.  (1899-1900.)  What  is  the  most  necessary  condition  for  the 
satisfactory  operation  of  a  pump?  What  is  the  advantage  of  a  suction 
chamber?  What  should  long  suction  pipes  be  provided  with? 

Ans.  69.  A  full  and  steady  supply  of  water.  The  pipe  connections 
should  in  no  case  be  smaller  than  the  openings  in  the  pump,  and  the 
suction  lift  and  delivery  pipe  should  be  as  straight  and  smooth  on  the 
inside  as  possible. 

A  suction  chamber  (air)  eliminates  pounding,  makes  the  action 
of  the  pump  easy  and  uniform,  also  facilitates  the  filling  of  the  barrel 
of  the  pump  when  at  high  speed. 

Long  suction  pipes  should  be  provided  with  a  foot-valve  just  above 
the  strainer  in  the  well  or  pit. 


Q-  70.  (1899-1900.)  For  boiler  feed  pumps,  or  pumps  doing  a  sim- 
ilar duty,  approximately  what  is  the  proportional  area  of  the  steam 
to  the  water  cylinder? 

Ans,  70.  The  steam  piston  should  have  about  2%  times  the  area 
of  the  water  piston.  There  being  no  mechanical  purchase  in  favor  of 

nvprttfiaam  P1S°n>  ^  mUSt  have  the  greater  area  of  the  tw°  in  order  to 
overbalance  the  pressure  on  the  water  piston 

138 


Q.  71.  (1899-1900.)  What  is  the  rule  for  finding  the  quantity  of 
water,  in  gallons,  pumped  in  one  minute,  at  100  piston  ft.  per  minute? 

Ans.  71.  The  diameter,  in  inches,  squared,  and  multiplied  by  4, 
equals  the  required  amount  of  water  in  gallons. 

4  D2  =  number  of  gallons. 


Q.  72.   (1899-1900.)     How  do  you  find  the  H.P.  necessary  to  pump 
water   to   a   given   height?     How   many   H.P.    are    required   to   pump 
(10,000,000)   ten  million  gallons  of  water  (150)  one  hundred  and  fifty 
ft.  high  in  (10)  ten  hours,  slippage  and  friction  not  taken  into  account? 
Ans.  72.     Multiply  the  total  weight  of  the  water  in  pounds  by  the 
height  in  feet,  and  divide  the  product  by  33000. 
10,000,000  gals,  pumped  in  10  hours. 
1,000,000  gals,  pumped  in  1  hour. 
16.666  gals,  pumped  in  1  minute. 
16.666  X  8  1/3  =  138,883  Ibs.  in  one  minute. 
138,883  X  150 

=  631.3  H.P. 

33,000 


Q.  73.  (1899-1900.)  What  have  you  to  say  in  regard  to  friction  and 
slippage  in  well  designed  and  constructed  pumps? 

Ans.  73.  In  well-designed  and  properly  constructed  steam  pumps 
the  slippage  will  amount  to  about  10  %  of  water  pumped,  and  an 
allowance  of  25  %  would  be  a  fair  allowance  for  both  the  slippage  and 
the  friction  of  the  pump. 

In  old  or  badly  constructed  pumps,  working  against  a  very  high 
pressure,  or  a  very  low.  lift,  the  net  loss  would  be  increased  to  twice 
the  percentage  given;  so  that,  in  calculating  the  size  of  a  feed  pump, 
allowing  for  leakage  of  water,  priming  of  the  steam,  blqwing  off  and  all 
other  leaks  that  may  occur.  For  a  steam  engine  the  pump  should  have 
the  capacity  of  2  to  2%  times  the  net  quantity  of  feed  water  required 
for  the  work  of  the  engine. 

For  marine  engines,  liable  to  use  salt  water,  the  size  of  the  pump 
should  be  so  as  to  be  able  to  deliver  from  3  to  4  times  as  much. 


Q.  74.  (1899-1900.)  From  what  is  power  developed  in  forcing  water 
into  a  boiler  by  an  injector?  Approximately,  what  is  the  difference  in 
the  velocity  of  the  steam  and  the  water? 

Ans.  74.  From  the  difference  in  the  velocity  of  the  escaping  steam 
from  a  boiler  under  pressure,  and  the  velocity  acquired  by  water  from 
the  same  boiler,  and  under  the  same  pressure,  and  at  the  same  time. 

Approximately,  the  steam  has  a  velocity  of  16  to  18  times  that  of 
the  water,  varying  with  the  different  pressures. 


Q.  76.  (1899-1900.)  Give  rule  for  finding  the  saving  which  may 
be  expected  by  heating  the  feed  water  a  given  amount. 

What  percentage  of  the  saving  of  fuel  may  be  expected  by  heating 
feed  water  from  60°  to  200°  for  a  boiler  carrying  125  Ibs.  gauge  pres- 
sure of  steam? 

Ans.  76.  Divide  the  difference  in  the  total  heat  of  the  water  above 
32°  F.,  before  and  after  heating,  by  the  total  heat  required  to  convert 
it  into  steam  from  the  given  initial  temperature;  multiply  this  quotient 
by  100,  and  the  product  will  be  the  percentage  of  saving  to  be  expected. 

Water  at  200°  contains  above  32°,  168.70  B.T.U. 

Water  at  60°  contains  above  32°,  28.01  B.T.U. 

139 


16g  70 28.01  =  140.69  =  the  number  of  heat  units  between  the  two 

temperatures  of  the  water  (60°  and  200°). 

There  is  in  steam,  at  125  gauge  pressure,  1189.5  B.T.U.  above  32° 
F  or  1189.5  —  28.01  =  1161.49  B.T.U.  above  60°  F. 

This  is  the  amount  of  heat  that  the  boiler  would  have  to  supply  to 
make  a  pound  of  steam  from  the  given  initial  temperature  of  60°  F.; 
but  we  have  seen  that  in  raising  the  water  to  200°  F.  140.69  units  are 
supplied  by  the  heater,  thus  saving  140.69  •*•  1161.49  of  the  heat  re- 
quired. 

140.69  X  100  -=-  1161.49  =  12.1  %,  the  percentage  of  saving  to  be  ex- 

The  same  result  may  be  found  from  the  use  of  the  table  showing 
the  increase  of  temperature  for  each  degree  of  initial  temperature  and 
steam  pressure;  thus,  in  the  column  of  60°  and  the  pressure  column 
of  125  Ibs.  we  find  the  increase  .0861,  the  saving  for  each  degree;  hence 
.0861  X  140  =  12.054  or  12.1%. 


Q.  100.  (1899-1900.)  With  an  engine  exhausting  into  a  surface 
condenser,  1,145  Ibs.  of  steam,  requiring  21,526  Ibs.  of  water  for  con- 
densation of  the  steam,  the  pressure  of  the  steam  entering  the  con- 
denser is  4  Ibs.  absolute  per  square  inch,  the  condensing  water  entering 
the  condenser  at  temperature  of  60°  F.,  discharging  at  115°  F.,  what 
was  the  temperature  of  the  condensation?  How  many  inches  of  vacuum 
was  being  maintained? 

Ans.  100.  Difference  in  temperature  of  the  entering  and  the  dis- 
charged water  equals  115°  F. —  60°  F.  =  553  F. 

The  B.  T.  U.  in  water  at  a  temperature  of  115°  F.  =  83.129.  The  B. 
T.  U.  in  water  at  a  temperature  of  60°  P.  =  28.009;  hence— 

The  number  of  heat  units  that  the  water  will  absorb  in  changing 
from  a  temperature  of  60°  F.  to  a  temperature  of  115°  F.  equals  83.129 
—  28.009  =  55.12  B.  T.  U. 

The  total  weight  of  the  water  used  21526  Ibs.  The  total  weight  of 
the  steam  condensed  1145  Ibs. 

Each  pound  of  steam  condensed  requires  as  many  pounds  of  water 
as  1145  is  contained  in  21526. 

21526  -T-  1145  =  18.80,  equals  the  number  of  pounds  of  water  required 
to  condense  one  pound  of  the  steam. 

One  pound  of  water  absorbs  55.12  B.  T.  U.  Then  18.80  Ibs.  water 
will  absorb  18.80  times  55.12  B.  T.  U. 

18.80  X  55.12  =  1036.256  B.  T.  U.  equals  the  number  of  heat  units 
absorbed  in  condensing  1  pound  of  the  steam. 

One  pound  of  steam  at  4  Ibs.  absolute  pressure  contains  a  total  heat 
of  1128.6  B.  T.  U.  above  32°  F.,  so  that  the  total  heat  equals  1128.6  added 
to  32,  1128.6  +  32  =  1160.6  B.  T.  U. 

If  the  water  absorbs  1036.256  B.  T.  U.  in  condensing  one  pound  of 
steam  then  it  must  leave  in  the  condenser  a  temperature  equal  to  the 
difference  between  1160.6  and  1036.256. 

1160.6  —  1036.256  =  124.344  B.  T.  U.  =  equiv.  temp.  124.04°  F.,  sus- 
taining a  vacuum  of  about  26.4  inches. 

In  practice  this  will  be  some  less  for  the  reason  that  the  air  gets 
into  the  condenser  and  the  pressure  will  be  higher  than  that  due  to 
the  temperature.  For  this  reason  it  is  common  to  see  condensers  with 
a  discharge  temperature  of  100°  F.,  which  by  the  tables  should  give  a 
vacuum  of  28  inches  and  over,  while  the  gage  shows  but  26  inches  and 
even  less. 


Q.  101.  (1899-1900.)  Describe  a  surface  condenser.  Describe  a  jet 
condenser.  Which  requires  the  greater  amount  of  water  for  condensa- 
tion purposes?  In  selecting  a  type  of  condenser  what  conditions  exist- 
ing would  govern  the  choice  between  the  two  types  of  condensers? 

140 


Ans.  101.  A  surface  condenser  is  a  vessel,  or  a  receptacle,  con- 
structed in  various  shapes,  with  double  heads  at  each  end. 

The  space  between  the  inner  heads  is  filled  with  small  brass  tubes 
varying  in  sizes  in  different  constructions  from  yz  inch  to  1  inch  diam- 
eter. 

In  some  constructions  the  tubes  are  made  fast  in  one  tube  sheet  and 
in  the  other  tube  sheet  made  secure  by  means  of  stuffing  box  and  gland. 

In  some  constructions  'they  are  made  secure  by  stuffing  box  and 
gland  at  both  ends. 

Some  constructions  the  tubes  are  made  fast  in  each  tube  sheet  and 
corrugated  their  entire  length. 

These  different  precautions  are  taken  for  an  allowance  of  expansion 
and  contraction  due  to  the  different  temperatures  to  which  they  are 
subjected. 

The  steam  is  usually  passed  through  the  annular  space  surrounding 
the  tubes  and  the  water  forced  through  the  tubes,  although  a  "vice 
versa"  process  is  allowable  and  is  sometimes  practiced. 

The  steam  from  the  exhaust  of  the  engine  coming  in  contact  with 
the  cool  surfaces  of  the  tubes,  is  condensed  and  falls  to  the  bottom  of 
the  condenser,  causing  a  partial  vacuum,  and  is  removed  by  the  means 
of  an  air  pump,  the  same  pump  also  performing  the  duty  of  removing 
any  air  that  may  enter  the  condenser. 

The  circulation  of  water  in  the  tubes  is  usually  kept  up  to  a  normal 
speed  per  minute  by  the  means  of  a  circulating  pump. 

Two  pumps  are  generally  used  with  a  surface  condenser. 

The  jet  condenser  is  also  constructed  in  various  shapes,  and  is 
generally  less  bulky  than  the  surface  condenser.  The  steam  from  the 
exhaust  pipe  of  engine  meeting  a  spray  of  cold  water,  is  condensed  and 
with  the  condensing  water  drops  to  the  bottom  of  the  condenser  and  is 
removed  by  the  aid  of  an  air  pump. 

This  form  of  a  condenser  requires  a  much  larger  air  pump  than 
the  surface  condenser,  the  air  pump  for  the  jet  condenser  having  to 
perform  the  double  duty  of  the  air  and  circulating  pump  for  the  sur- 
face condenser. 

The  surface  condenser  requires  a  much  larger  quantity  of  condens- 
ing water  than  the  jet  condenser. 

The  type  of  condenser  to  select  for  installation  in  a  plant  depends 
upon  a  great  many  local  circumstances,  a  few  of  which  are:  Space 
available,  condition  or  quality  of  water  to  be  used  for  condensation 
purposes;  also  the  cost  of  installation. 


Q.  102.  (1899-1900.)  What  relative  duty  does  an  air-pump  perform 
for  a  condenser? 

What  relative  duty  does  a  circulating-pump  perform  for  a  surface 
condenser? 

Give  rule,  or  express  in  formula,  for  finding  the  diameter  of  a  sin- 
gle acting  air-pump  for  a  jet  condenser  (using  a  stroke  of  18  inches). 

Ans.  102.  The  relative  duty  of  an  air  pump  to  a  condenser  is  to 
remove  the  water  of  condensation  and  the  air  that  enters  the  con- 
denser with  the  steam,  and  through  minor  leaks. 

For  a  jet  condenser,  in  addition  to  the  above  enumerated  duties,  it 
has  to  remove  the  condensing  water. 

Rule. — Multiply  the  total  number  of  pounds  to  be  condensed  per 
minute  by  the  number  of  pounds  of  cooling  water  per  pound  of  steam; 
reduce  the  pounds  of  cooling  water  used  per  minute  to  cubic  feet.  Also 
the  water  of  condensation;  add  the  two  together  and  their  sum  equals 
the  volume  of  water  to  be  handled  per  minute.  Divide  this  volume  of 
water  by  the  number  of  strokes  the  pump  makes  multiplied  by  the 
length  of  the  stroke  and  the  quotient  equals  the  cross-sectional  area 

141 


of  the  cylinder  of  the  pump;  theoretically,  for  making  allowance  for 
the  air  to  be  removed,  multiply  the  volume  of  water  by  2.75  and  the 
product  is  the  area  of  the  piston  required. 

When  the  air  and  water  are  removed  a  partial  vacuum  is  formed 
in  the  condenser  causing  the  condensing  water  to  be  lifted  into  it  by 
atmospheric  pressure  if  the  vertical  distance  is  not  above  20  feet. 
•  The  relative  duty  of  a  circulating  pump  to  a  surface  condenser  is 
to  circulate  the  condensing  water  through  the  tubes  of  the  condenser  to 
absorb  the  heat  in  the  steam. 

In  addition  to  the  volume  of  water  to  be  removed  there  is  a  large 
quantity  of  air;  therefore  the  displacement  in  the  air  pump  for  a  jet 
condenser  must  be  in  excess  of  the  volume  of  water  to  be  displaced. 

Good  authorities  place  this  at  2.75  times  the  volume  of  water  for 
single  acting  pumps. 

Formula: 

For  the  volume  of  single  acting  pumps — 
Q  +  q 

Volume  =  2.75 

n 

In  which— 

Q  —  volume  of  condensing  water. 

q  =  volume  of  water  for  condensation. 

n  =  number  of  strokes  of  pump  per  minute. 

Q.  103.  (1899-1900.)  Find  the  diameter  of  a  single-acting  air-pump 
for  a  condenser  to  an  engine  developing  500  I.  H.  P.  using  18  Ibs.  of 
steam  per  I.  H.  P..  per  hour,  exhausting  into  a  condenser  (jet)  at  a 
temperature  of  140°  F.,  using  22  Ibs.  of  water  for  condensing  purposes 
per  pound  of  steam  used;  the  temperature  of  the  condensing  water 
60°  F.;  air  pump  with  a  stroke  of  18  inches,  making  140  strokes  per 
minute. 

Ans.  103.  500  X  18  =  9000  Ibs.  of  steam  to  be  condensed  per  hour 
=  9000  -5-  60  =  150  Ibs.  per  minute;  150  X  22  =  3300  Ibs.  of  condensing 
water  used  per  minute. 

3300  Ibs.  of  water  at  60°  F.,  volume=  52.91  c.  ft. 

150  Ibs.  of  water  at  140°  F.,  volume  =  2.44  c.  ft. 

Total  volume  of  water  to  be  moved  per  minute  by  air  pump  =  52.91 
+  2:44  =  55.35  c.  ft.  55.35  H- 70  (strokes  of  pump)  =.79  c.  ft.  per 
stroke.  .79  c.  ft.  X  1728  =  1365.12  c.  in.  =  volume  of  water  per  stroke 
of  pump. 

1365.12  X  2.75 

=  208.56  inches,  the  area  of  the  cross-section  of  the 

18 
pump  cylinder. 

/20O6 
^/  — =  16.3  inches,  the  diameter  of  the  cylinder. 

Or,  16"  X  18"  pump  at  70  strokes  per  minute. 


Q.  104.  (1899-1900.)  Upon  what  does  the  amount  of  condensing 
water  depend?  Give  the  rule,  or  express  the  formula,  for  finding  the 
amount  of  water  required. 

What  quantity  of  condensing  water  per  pound  of  steam  is  required 
to  condense  the  steam  exhausting  into  a  condenser  at  140°  F  •  the 
temperature  of  the  condensing  water  60°  F.,  and  the  temperature  of 
t&e  hot-well  110°  F. 

Ans.  104.  The  amount  of  condensing  water  depends  upon  its  tem- 
perature entering  the  condenser,  and  the  amount  of  heat  to  be  absorbed 

142 


from  the  steam;  also  upon  the  type  of  condenser  used;  the  jet  con- 
denser taking  less  water  than  the  surface  condenser  to  perform  the 
same  work;  the  quantity  depends  principally  upon  the  difference  of 
temperature  of  the  water  and  the  steam. 

Rule. — Subtract  from  the  heat  of  the  steam  at  the  terminal  pres- 
sure, the  heat  in  the  water  of  condensation  as  it  leaves  the  condenser, 
and  divide  this  remainder  by  the  difference  in  the  temperatures  of  the 
water  before  and  after  passing  through  the  condenser. 
Formula — 

HW 

Q  

R 

In  which — 

Q  —  quantity  of  cooling  water. 

H  =  heat  units  given  up  by  the  steam  in  condensing. 

W  =  weight   in   pounds  of   steam    condensed. 
R  =  difference  in  temperature  of  cooling  water  and  that  of  hot  well. 

There  must  be  sufficient  water  mixed  with  the  steam  to  absorb  and 
reduce  the  heat  of  the  steam  at  a  temperature  of  140°  F.  to  110°  F.,  the 
temperature  of  the  hot  well. 

The  total  heat  of  steam  at  140°  F.  above  32°  F.  =  1124.64  B.  T.  U. 

The  total  heat  of  water  from  the  condenser  at  110°  F.  above  32°  F. 
equals  78.11  B.  T.  U. 

1124.64  —  78.11  =  1046.53  B.  T.  U.  absorbed  by  the  cooling  water. 

The  difference  in  temperature  of  the  cooling  water  and  the  dis- 
charge from  the  condenser  equals  110°  —  60°  =  50°  F. 

Then  1046.53  -=-  50  =  20.93  Ibs.  water  to  condense  one  pound  of 
steam  at  a  temperature  of  140°  F.  to  110°  F. 


143 


Electricity,  Dynamos,  Motors,  Wiring, 


Etc. 


Q.  60.  (1896-7.)  What  is  the  difference  between  a  kilowatt  and  an 
electrical  horse  power? 

Ans.  60.  One  electrical  horse-power  equals  746  watts;  one  kilowatt 
equals  1,000  watts. 

Then  1  kilowatt  equals  1,000  divided  by  746,  or  1.3405  H.P.  Or, 
again,  a  kilowatt  is  1,000  watts.  An  electrical  horse-power  is  746  watts; 
therefore  the  difference  is  254  watts. 


Q.  61.     (1897-8.)     (a)  What  is  a  "volt"? 

(b)    How  do  you  fix  its  value  in  your  mind  for  practical  purposes? 

Ans.  61.  (a)  The  volt  is  the  practical  unit  of  electro-motive  force, 
electrical  pressure  (head)  or  difference  of  potential.  It  is  equal  to  10s 
absolute,  or  C.G.S.  units.  An  electro-motive  force  of  one  volt  will  pro- 
duce a  current  of  one  ampere  when  applied  to  a  conductor  having  a 
resistance  of  one  ohm. 

(b)  Practically  one  cell  of  ordinary  battery  gives  an  electro-motive 
force  of  one  volt.  Latimer  Clark's  standard  cell  gives  1.436  volts  under 
favorable  conditions. 

The  volt  is  usually  conceived  as  about  the  E.  M.  F.  of  a  Daniels  cell, 
or  the  ordinary  voltaic  cell.  This  last  is  somewhat  inaccurate,  however, 
as  the  different  cells  have  different  pressures,  from  between  .6  and  .7 
in  the  Edison  Lalande  to  nearly  two  volts  in  the  Grove,  Bunsen,  Grenet, 
etc. 

One  association  defines  the  volt  as  the  pressure  induced  in  a  con- 
ductor [one  foot  in  length]  passing  through  a  magnetic  field  at  such 
a  speed  as  to  cut  the  lines  of  force  at  the  rate  of  100,000,000  lines  per 
second.  This  is  a  useful  conception  in  dealing  with  dynamos,  but 
would  seem  more  real  to  the  writer  if  "a  field  capable  of  an  aggregate 
pull  of  225  Ibs."  could  be  substituted  for  a  field  of  100,000,000  lines  of 
force. 

An  idea  of  the  strength  of  110  volts  may  be  obtained  by  putting  the 
end  of  your  finger  in  a  lamp  socket  having  this  potential. 


Q.  62.     (1897-8.)     What  is  an  "ohm"? 

(b)  What  is  the  resistance  of  one  hundred  feet  of  No.  19  American 
gage  copper  wire? 

Ans.  62.  (a)  The  ohm  is  the  practical  unit  of  electrical  resistance. 
It  is  equal  to  10"  absolute,  or  C.G.S.  units  of  resistance.  The  legal  ohm 
is  the  resistance  of  a  column  of  mercury  1  sq.  millimeter  in  section, 
and  106  centimeters  long,  at  a  temperature  of  32°  F. 

We  have  thought  that  the  ohm  was  best  conceived  of,  as  the  resis- 
tance of  a  certain  length  of  a  certain  sized  wire.  The  writer  always 
thinks  of  it  as  the  resistance  of  a  certain  piece  of  german  silver  wire 
that  he  has  strung  on  a  board  between  binding-posts  for  the  purpose  of 
estimating  the  resistance  (internal)  of  voltaic  cells,  etc. 

144 


An  idea  of  the  resistance  of  one  ohm  may  be  had  from  the  follow- 
ing table,  which  gives  the  number  of  feet  and  number  (B.  &  S.)  of  cop- 
per wire  having  a  resistance  of  one  ohm: 

FEET  PEB  OHM   OF  WIRE    (B.   &   S.) 

94  feet  of  No.  20. 


150 

239 

380 

605 

961 

1529 

2432 

3867 


18. 
16. 
14. 
12. 
10. 

8. 

6. 

4. 


(b)  We  find  in  Kent's  tables  that  124.4  ft.  of  No.  19  A.W.G.  copper 
wire  has  a  resistance  of  1  ohm.  Therefore  1  foot  has  a  resistance  of 
.008038  ohm;  100  feet  a  resistance  of  .8038  ohm. 


Q.  63.     (1897-8.)     What  is  an  ampere? 

(b)     How  do  you  fix  its  value  in  your  mind  for  practical  purposes? 

Ans.  63.  (a)  The  ampere  is  a  certain  quantity  of  electricity  flowing 
in  a  certain  time.  It  is  the  practical  unit  of  current.  It  is  one-tenth 
of  the  C.G.S.  unit  of  current.  The  C.G.S.  unit  of  current  is  that  current 
which,  flowing  through  a  length  of  one  centimeter  of  wire,  acts  with  a 
force  of  one  dyne  upon  a  unit  of  magnetism  distant  one  centimeter  from 
every  point  of  the  wire. 

(This  C.G.S.  definition  is  a  good  illustration  of  the  un-get-at-ableness 
of  electrical  definitions.  The  unit  quantity  of  magnetism,  or  unit  pole, 
is  an  imaginary  thing  that  does  not  and  cannot  exist.  The  centimeter 
is  about  .4"  and  the  dyne  about  1/445,000  of  a  pound,  however.  Is  there, 
then,  no  iconoclastic  balm  in  Gilead  that  will  mitigate  this  apparently 
necessary  evil? — Ed.  Com.) 

(b)  An  ordinary  55-volt  16  C.P.  lamp  requires  a  current  of  one  am- 
pere; a  110-volt  lamp,  .5  ampere;  the  ordinary  2000  C.P.  arc  circuit 
requires  9.6  amperes,  or  practically  one  C.G.S.  unit. 


Q.  64.     (1897-8.)     What  is  a  watt? 

(b)  What  is  its  value  in  foot-lbs.  per  second,  and  in  B.T.U.  per 
second? 

Ans.  64.  (a)  The  watt  is  the  practical  unit  of  electrical  power.  It 
is  equal  to  10,000,000  units  of  power  in  the  C.G.S.  system.  To  measure 
the  power  carried  by  an  electrical  circuit  we  must  measure  both  the 
volts  and  the  amperes.  The  product  of  the  volts  and  amperes  is  the 
watts,  or  electrical  power:  746  watts  —  one  horse-power. 

(b)     The  value  of  a  watt  in  foot-pounds  per  sec.  is 

33,000 

=.7373  ft.  Ibs. 

60  X  746 

The  value  of  a  watt  in  B.T.U.  per  sec.  is  .7373  -h  778  =  .0009477  H.U. 
per  sec. 

We  have  thought  that  the  watt  was  best  thought  of  in  reference  to 
its  value  in  foot-pounds  and  in  heat  units.  It  is  said  that  a  wire  will 
throw  off  about  .005122  heat  units  per  square  foot  of  surface  and  per 
second  for  each  degree  F.,  that  its  temperature  is  above  that  of  the 
surrounding  atmosphere.  Knowing  the  square  feet  of  surface  of  the 
wire,  the  number  of  watts  expended  in  it  and  the  value  of  a  watt  in 
heat  units,  it  is  easy  to  calculate  how  hot  the  wire  will  become,  and 
how  much  of  the  power  is  wasted  in  heat. 

]45 


Q.  £5.     (1897-8.)     Wliat  is  Ohm's  law? 
Ans.  65.     Ohm's  law:  — 

(1)  Amperes  =  volts  -r-  ohms  =  A  =  — 

( 2 )  Volts  =  amperes  X  ohms  =V  =  AO 

(3)  Ohms  =  volts :-=-  amperes  =  O  =  — 

A 

we  will  have  a  1000-volt  circuit,  or  the  difference  of  potential  is  said  to 
be  1000  volts.  The  potential  decreases  along  the  circuit  in  proportion 
to  the  resistance.  We  would  say,  for  instance,  the  positive  terminal 
of  the  dynamo  has  1000  volts  potential,  the  negative  terminal  has  zero 
potential.  After  the  current  has  passed  ten  lamps  the  difference  of 
potential  is  500  volts,  etc. 

The  potential  of  the  earth  is  zero,  and  if  a  connection  with  the  earth 
and  the  circuit  is  made  at  any  point  the  current  runs  to  earth,  or  is 
"grounded." 


Q.  66.  (1897-8.)  What  is  meant  by  the  difference  of  potential  of  a 
current? 

Ans.  66.  The  potential  of  the  current  refers  to  the  voltage  between 
the  terminals  of  the  circuit,  generally  speaking.  If  we  take,  for  exam- 
ple, ^n  arc  light  circuit  supplying  20  lamps  requiring  50  volts  per  lamp, 


Q;  67.  (1897-8.)  What  are  the  advantages  and  disadvantages  of 
high  tension  currents? 

Ans.  67.  The  advantages  of  high  potential  currents  consist  in  the 
ability  to  transmit  electricity  cheaply,  as  a  smaller  wire  is  required  to 
transmit  a  given  amount  of  electricity  at  a  higher  potential  than  at  a 
lower  one.  Small  wires  cost  less  than  large  ones. 

•  The  disadvantages  of  high  potential  currents  arise  from  the  difficulty 
of  insulation,  requiring  more  expensive  insulators.  High  potentials  are 
also  dangerous  to  human  life. 


--  Q.  68.     (1897-8.)     What  is  a  transformer  and  how  constructed? 

Ans.  68.  A  transformer  is  an  apparatus  for  transforming  currents 
of  high  potential  and  small  amperage  into  currents  of  low  potential  and 
large  amperage,  and  vice  versa. 

The  kind  commonly  used  for  transforming  alternating  currents  con- 
sists of  a  core  of  laminated  iron  plates  with  two  sets  of  windings.  The 
primary  current  is  sent  by  the  dynamo  through  the  primary  coil  and 
an  induced  current  is  generated  thereby  in  the  secondary  coil.  The 
change  in  potential  and  amperage  is  regulated  by  the  size  of  wire  and 
number  of  turns  in  the  different  windings. 


Q.  69.     (1897-8.)     Can  a  direct  current  be  transformed?    If  so,  how? 

Ans.  69.  A  direct  current  can  be  transformed  by  means  of  a  rotary 
transformer.  This  consists  of  a  motor  having  two  sets  of  windings 
and  two  commutators.  The  primary  current  is  sent  through  one  wind- 
ihg  and  the  armature  turns,  generating  a  secondary  current  in  the  other 
windings.  By  properly  proportioning  the  windings  a  current  of  the 

ired  voltage  can  be  obtained  from  the  secondary.  These  machines 
are  in  common  use  in  large  central  stations,  where  they  are  commonly 
termed  "boosters."  Stationary  or  static  transformers  are  used  for 
alternating  currents. 

146 


Q.  70.  (1897-8.)  In  a  circuit  there  is  a  shunt  that  has  10  times  as 
much  resistance  as  the  part  of  the  main  circuit  between  the  terminals 
of  the  shunt,  what  part  of  the  total  current  will  go  through  the  shunt? 

Ans.  70.  In  dividing  a  circuit  the  current  divides  inversely  propor- 
tional to  the  resistance  of  the  circuits. 

In  case  the  shunted  circuit  has  ten  times  the  resistance  of  that  part 
of  the  main  circuit  between  the  terminal  of  the  shunt,  there  will  be  one- 
tenth  as  much  current  through  the  shunt  as  through  the  main  connec- 
tion between  the  terminals  of  the  shunt.  We  may  say  that  the  current 
is  therefore  divided  into  eleven  parts,  one-eleventh  going  through  the 
shunt  and  ten-elevenths  going  through  the  main  circuit. 

Where  the  main  current  flowing  in  the  main  circuit  is  too  large  to 
be  conveniently  measured,  the  ammeter  is  frequently  placed  in  a  shunt 
through  which  a  certain  definite  portion  of  the  current  is  known  to  be 
flowing. 

•  Q.  71.  (1897-8.)  What  is  meant  by  the  carrying  capacity  of  a  wire? 
Ans.  71.  By  carrying  capacity  of  a  wire  is  meant  the  amount  of 
current  it  will  carry  without  heating  to  such  an  extent  as  to  increase 
the  resistance  too  much  or  to  cause  danger  of  burning  the  insulation 
or  wood  work. 

Q.  72.  (1897-8.)  What  is  the  allowable  carrying  capacity  of  the 
usual  sizes  of  wire  from  No.  0000  to  No.  18  B.  and  S.  gage — according  to 
the  National  Board  of  Underwriters? 

Ans.  72.  The  rules  of  the  National  Board  of  Underwriters  (1897) 
gave  the  following  table  of  carrying  capacities: 


No.  of  wire, 
B.&S.  gage. 

18 

16 

14 

12 

10 

8 

6 

5 

4 

3   . 

2 

1 

0 

00 

000 

0000 


Rubber- 

Circular 

covered 

Mils. 

Wire. 

Amperes. 

1624 

3 

2583 

6 

4107 

12 

6530 

17 

10382 

24 

16510 

33 

26250 

46 

33102 

54 

41743 

65 

52634 

76 

66373 

90 

83694 

107 

105593 

129 

133079 

150 

167805" 

177 

211600 

210 

Weather- 
proof 
Wire. 

Amperes. 
5 
8 

16 
23 
32 
40 
64 
77 
92 
110 
131 
156 
185 
220 
262 
312 


Q.  73.     (1897-8.)     How  much  current  is  used  by  a  16  c-p  lamp  on  an 
average? 

Ans.  73.    A  55-volt  16  C.P.  lamp  requires  one  ampere. 

A  110-volt  16  C.P.  lamp  requires  .5  ampere,  on  an  average. 


Q.  74.     (1897-8.)     What  is  meant  by  the  "drop"  in  a  circuit? 

Ans.  74.  The  drop  in  a  circuit  means  the  number  of  volts  lost  in 
overcoming  the  resistance  of  the  circuit.  It  corresponds  to  the  drop 
in  pressure  along  a  steam  pipe  or  water  main;  in  the  latter  case  some- 
times called  "loss  of  head." 


147 


Q.  75.  (1897-8.)  What  is  the  allowable  drop  (a)  for  electric  light 
wiring,  (b)  for  wiring  to  motors  for  power? 

Ans.  75.  In  electrical  wiring  for  lights  various  percentages  of  drop 
are  allowed,  from  1%  to  10%,  depending  on  circumstances.  The  larger 
per  cent  is  used  when  the  distance  is  long,  as  it  is  then  cheaper  to 
generate  the  extra  electricity  than  it  is  to  put  in  large  wires. 

In  wiring  for  motors  various  percentages  of  drop  are  used,  from  1% 
to  30%,  according  to  circumstances.  It  often  happens  that  there  is  as 
much  as  30%  drop  in  street  railway  circuits  at  the  end  of  long  lines. 

In  ordinary  shop  practice  about  3%  drop  is  allowed  for  lights  and 
5%  for  motors. 

Q.  76.  (1897-8.)  How  can.  the  size  of  wire  to  be  used  be  determined 
for  conveying  a  given  number  of  amperes  a  given  distance  with  a  given 
drop? 

Ans.  76.  A  wire  which  conveys  a  current  of  electricity  must  have 
an  area  of  as  many  circular  mils  in  its  cross-section  as  is  equal  to  21 
times  the  distance  transmitted  in  feet,  multiplied  by  the  current  in 
amperes  and  divided  by  the  drop  in  volts. 

The  constant  21  equals  the  resistance,  in  ohms,  of  a  wire  one  circular 
mil  in  diameter  and  2  ft.  in  length.  By  taking  the  resistance  of  2  ft. 
length  in  this  calculation  it  is  then  necessary  to  take  only  the  distance 
in  feet  one  way  in  the  calculation. 

21  X  Amperes  X  Dist.  in  ft. 

Area  in  Cir.  Mils  = 

Drop  in  volts. 

After  finding  the  circular  mils,  look  up  the  corresponding  size  of 
wire  in  a  table.  (See  Ans.  No.  72.) 

Example: — 150  amperes  are  carried  150  feet  with  3%  drop,  120  volts 
pressure.  By  the  rule:  21  X  150  X  150  =  472,500.  The  number  of  volts 
drop  is  120  X  .03  =  3.6;  then  472,500  H-  3.6  =  131,233  mils.  Turning  to 
answer  No.  72,  we  find  that  this  corresponds  to  00  wire,  very  nearly. 


Q.  77.  (1897-8.)  What  is  a  dynamo?  How  does  it  generate  elec- 
tricity? 

Ans.  77.  A  dynamo  is  a  machine  for  converting  energy,  in  the  form 
of  mechanical  power,  into  energy  in  the  form  of  an  electric  current,  or 
vice  versa,  by  the  operation  of  setting  conductors,  usually  coils  of  copper 
wire,  to  rotate  in  a  magnetic  field,  or  by  varying  a  magnetic  field  in 
the  presence  of  conductors.  A  dynamo  consists  of  two  essential  parts, 
a  field-magnet  and  an  armature.  The  field-magnet  must  be  magnetized 
either  by  the  dynamo  itself  or  some  outside  source.  The  armature 
consists  of  coils  of  insulated  copper  wire  wound  upon  a  core  which  is 
mounted  on  a  shaft  and  caused  to  rotate  between  the  poles  of  the  field- 
magnet  in  such  a  manner  that  the  lines  of  magnetic  force  passing 
through  the  coils  of  the  armature  are  constantly  increasing  or  decreas- 
ing. This  action  causes  currents  of  electricity  to  be  generated  in  the 
coils  of  the  armature,  which  are  collected  from  rings  or  a  commutator 
by  suitable  brushes  and  sent  out  on  the  circuit  through  suitable  con- 
ductors. 


Q.  78.     (1897-8.)    What  is  meant  by  the  magnetic  circuit? 

Ans.  78.  The  magnetic  circuit  is  the  path  or  space  through  which 
the  magnetic  lines  or  force  travel. 

Take,  for  example,  the  Edison  dynamo.  The  field-magnets  of  this 
machine  consist  of  five  pieces,  viz.,  the  pole  pieces  D,  E;  cores  B,  C,  and 
a  yoke,  A.  (Fig.  8.)  When  the  magnet  is  in  action  the  strongest  field 
is  between  the  poles,  D  and  E,  but  the  magnetic  circuit  follows  around 
through  all  the  parts,  D,  B,  A,  C,  E,  and  across  to  D.  If  an  armature 

148 


FIG.  8. 

with  an  iron  core  be  placed  in  the  space  between  the  pole-pieces,  the 
magnetic  circuit  between  the  pole-pieces  is  intensified,  as  magnetism 
travels  through  iron  easier  than  through  air. 

(Note:— In  the  last  sentence  we  would  say  that  with  the  same 
impelling  force  [ampere-turns]  more  magnetism  would  pass  through  the 
circuit  because  the  resistance  was  less.— Ed.  Com.  1897-8.) 


Q.  79.  (1897-8.)  What  is  the  relative  permeability  to  magnetism 
of  iron  and  air? 

Ans.  79.    S.  P.  Thompson's  third  edition,  pages  55  and  56,  says:  — 

"If  we  take  the  magnetic  permeability  of  air  as  1,  then  the  permea- 
bility of  iron  will  be  represented  by  values  lying  between  5  and  20,000, 
according  to  the  quality  of  the  iron"  [and  the  density  of  the  lines  of 
force]. 

The  conception  of  the  magnetic  circuit  and  its  close  analogy  to  the 
electric  circuit  is  comparatively  modern  and  exceedingly  useful.  One 
can  think  of  the  air  gaps  as  affording  thousands  of  times  as  much 
resistance  as  iron  of  equal  length  and  cross-section. 

In  Kent's  handbook  we  find  values  of  permeability  showing  that  for 
grey  cast  iron  it  is  from  37  to  800  times  that  of  air,  and  that  the  perme- 
ability of  annealed  wrought  iron  is  from  54  to  25,000  times  that  of  air. 


Q.  80. 
currents. 

Ans.  8 


(1897-8.)     Give  a  table  of  diameters  for  fuses  for  different 


The  diameter  of  a  fuse  may  be  calculated  as  follows:  — 
a=(c-£-a)f 

in  which  d  is  the  diameter  in  inches  of  fuse,  c  the  number  of  amperes 

carried,  and  a,  a  constant  which  varies  according  to  the  material  of 

which  the  fuse  is  composed. 

The  following  are  values  of  the  constant,  a,  for  the  metals  named: 

Copper  =  10224;    tin  =  1642;    lead  =  1379;    aluminum  =  7585;    iron  = 

3150. 

From  the  above  formula  may  be  deduced  the  following: 

Rule: — Divide  the  constant,   for  the  given   metal,  by  the  number 

of  amperes  required;  multiply  this  quotient  by  itself   (square  it)   and 

149 


extract  the  cube  root  of  this  product  (square).    Call  this  result  No.  1. 

Second.    Take  unity  (one)  as  the  numerator  and  result  No.  1  as  the 
denominator  of  a  fraction. 

This  fraction  expresses  in  inches  the  diameter  required. 

RUPTURING  CURRENT  OF  FUSES. 


n  M    Tin  Wire 

"c  £ 
ft 

11 

go 

u 

SM 

<im 

1 

.0072 

36 

2 

.0113 

31 

3 

.0149 

28 

4 

.0181 

26 

5 

.0210 

25 

10 

.0334 

21 

15 

.0437 

19 

20 

.0529 

17 

25 

.0614 

16 

30 

.0694 

15 

35 

.0769 

14.5 

40 

.0840 

13.5 

45 

.0909 

13 

50 

.0975 

12.5 

60 

.1101 

11 

70 

.1220 

10 

80 

.1334 

9.5 

90 

.1443 

9 

100 

.1548 

8.5 

120 

.1748 

7 

140 

.1937 

6 

160 

.2-118 

5 

180 

.2291 

4 

200 

.2457 

3.5 

250 

.2851 

1.5 

300 

.3220 

0 

Lead  Wire 

|s  S<? 

ii  m 

.0081  35 

.0128  30 

.0168  27 

.0203  25 

.0236  23 

.0375  20 

.0491  18 

.0595  17 


Copper  Wire 

£"5   ks* 
.0021   47 


.0690   15 
.0779   14 
.0864   13.5 
.0944   13 
.1021   12 
.1095   11.5 
.1237   10.5 
.1371   10 
.1499 
.1621 
.1739 
.1964 
.2176 
.2379 


9.5 


.2573 
.2760 
.3203 
.3617 


7 
6 
5 
4 
3 
2 
0 
0.5 


.0034  43 
.0044  41 
.0053  39 
.0062  38 
.0098  33 
.0129  30 
.0156  28 
.0181  26 
.0205  25 
.0227  24 
.0248  23 
.0268  22 
.0288  22 
.0325  21 
.0360  20 
.0394  19 
.0426  18.5 
.0457  18 
17.5 
17 
16 
16 

.0725  15 
.0841  13.5 
.0950  12.5 


.0516 
.0572 
.0625 
.0676 


5 

<^ 

<  ° 

.0047 

40 

.0026 

.0074 

36 

.004] 

.0097 

33 

.0054 

.0117 

31 

.0065 

.0136 

29 

.0076 

.0216 

24 

.0120 

.0283 

22 

.0158 

.0343 

20.5 

.0191 

.0398 

19 

.0222 

.0450 

18.5 

.0250 

.0498 

18 

.0277 

.0545 

17 

.0308 

.0598 

16.5 

.0328 

.0632 

16 

.0352 

.0714 

15 

.0397 

.0791 

14 

.0440 

.0864 

13.5 

.0481 

.0935 

13 

.0520 

.1003 

12 

.0558 

.1133 

11 

.0630 

.1255 

10 

.0698 

.1372 

9.5 

.0763 

.1484 

9 

.0826 

.1592 

8 

.0886 

.1848 

6.5 

.1028 

.2086 

5 

.1161 

Con.  1642.        Con.  1379.        Con.  10244.        Con.  3150.        Con.  7585 

Omaha  No.  1,  Nebraska. 

Table  from  Kent's  M.  E.  Handbook,  2d  Ed.       Per  J.  A.  Bramhall,  Inst. 
(Ex.  Q.     What  is  the  diameter  required  for  a  lead  fuse  to  carry 
thirty  (30)  amperes  of  current? 

The  constant  for  lead  is  1379;  this  divided  by  30  is  a  little  less  than 
46;  46  multiplied  by  itself  (square)  is  2116;  the  cube  root  of  2116  is  a 
little  greater  than  12.8. 

Therefore  the  fraction  is: 

1 

-  =  .078 
12.8 
which  is  the  diameter  required,  approximately. 

If  one  wishes  to  obtain  the  constant  for  a  given  metal  or  alloy  he 
may  do  so  by  the  following  formula: 


Vds 

Example:    A  tin  wire  .021"  diam.  melts  with  a  current  of  5  amperes, 
what  is  the  constant  for  tin? 
c  =  5 

d  =  .021  .-.  ds  =  . 000009261 
Vd3  =  V.000009261  =  .003044. 

c               5             5000000 
Therefore  constant  —  a  = = = =  1642 

Vd8      .003044          3044 
150 


In  reference  to  Q.  80  some  of  the  associations  have  thought  that  no 
useful  table  could  be  constructed  because  the  material  of  the  wire 
varied  and  the  result  was  influenced  by  the  length  of  the  fuse.  It 
seems  to  us  that  the  only  care  in  reference  to  the  length  of  the  fuse 
should  be,  that  the  fuse  should  not  be  so  short  that  its  points  of  attach- 
ment will  have  an  appreciable  effect  in  conducting  away  and  radiating 
the  heat  generated  in  the  fuse.  We  think  the  table  and  formula  given 
should  prove  useful.  

Q.  30.  (1898-9.)  Give  the  principal  Electrical  Units  which  the  en- 
gineer should  be  conversant  with. 

Ans.  30.  The  principal  electrical  units  which  the  engineer  should 
be  familiar  with  are  as  follows: 

The  "Ohm" — unit  of  resistance;  the  resistance  of  conductor  passing 
one  "ampere"  of  current,  when  the  electro-motive  or  impelling  force  is 
one  "volt." 

The  "Ampere" — unit  of  current  strength;  the  quantity  of  current 
flowing  through  the  one  "ohm"  or  resistance,  when  the  electro-motive 
force  is  one  "volt." 

The  Volt — unit  of  electro-motive  force;  the  impelling  force  or 
"electrical  pressure"  required  to  cause  "flow"  of  one  "ampere"  through 
a  conductor  offering  one  ohm  resistance. 

The  Watt — unit  of  power;  equivalent  to  1-746  HP.  The  Kilowatt — 
1,000  watts. 


PREFACE  TO  QUESTIONS  63  ET  SEQ.  (1  98-9). 

It  is  a  rule,  rather  than  the  exception,  wheresoever  steam  engines 
and  electrical  generators  are  jointly  placed  in  any  plant,  that  the  care 
and  supervision  of  both  is  assigned  to  one  "head."  Such  installations 
having  become  so  common,  the  stationary  engineer  can  no  longer  afford 
to  be  deficient  in  knowledge  of  the  practical  points  involved  in  a  matter 
which  is  universally  regarded  as  a  prime  necessity,  and  which,  by 
common  consent,  Seems  to  have  become  incorporated  with  his  other 
multifarious  duties. 

With  the  steam  engine  as  the  prime  mover,  the  dynamo,  with  all 
usually  appurtenant  thereto,  as  the  means  for  transposing  the  power 
developed,  we  have  a  combination  which  is  quite  likely  to  remain  sub- 
stantially and  inseparably  linked  together;  hence  under  the  general 
condition  of  affairs,  no  engineer  to-day  can  regard  himself  thoroughly 
equipped  in  his  calling  unless  he  has  acquired  a  more  than  fairly  good 
conception  of  the  current  practice  which  befits  the  case. 

There  can  be  no  error,  therefore,  in  including,  under  the  head  Of 
"Practical  Steam  Engineering,"  such  questions  as  are  pertinent  to 
electrical  installations.  While  this  subject  in  itself  is  too  vast  to  per- 
mit of  exhaustive  treatment,  yet  it  is  hoped  that  the  nature  of  the 
queries  will  afford  timely  discussion  and  thereby  serve  the  purpose  of 
association  instruction.  It  will  be  noted  that  some  very  leading 
questions  are  asked  and  satisfactory  answers  cannot  fail  to  interest 
those  who  are  just  "breaking  in"  on  electrical  matters,  or  who  are  likely 
to  be  confronted  with  similar  propositions. 

The  units  pertaining  to  electricity  are  of  primary  importance,  and 
while  these  are  not  specifically  embraced  amongst  the  questions,  it  is 
suggested  that  their  definitions  be  reviewed  during  association  discus- 
sion, in  order  that  the  uninformed  may  be  started  with  a  proper  under- 
standing of  their  nature,  their  values  and  their  application. 


Q.  63.  (1898-9.)  What  voltage  do  you  consider  best  adapted  for  a 
factory  plant,  where  motors,  incandescent  and  arc  lamps  are  served 
from  one  dynamo,  the  aggregate  work  approximating  100  HP?  Give 
reasons. 


Ans.  63.  For  a  factory  plant  where  motors,  incandescent  and  arc 
lamps  are  served  from  one  dynamo,  a  voltage  of  110  would  be  the  best. 
However,  if  lamp  manufacturers  would  perfect  a  reliable  220  volt  lamp 
this  voltage  for  such  work  would  have  its  advantages. 

The  advantage  of  110  volts  for  an  installation  of  this  kind  are  safety 
and  simplicity. 

The  disadvantages  are  higher  expenses  of  copper  for  the  power 
developed. 

Q.  64.  (1898-9.)  Which  is  preferable  for  the  conditions  noted  in 
Q.  63,  a  shunt  or  compound  wound  dynamo?  How  do  these  differ  from 
"series"  machines? 

Ans.  64.  A  compound  wound  dynamo  is  the  only  satisfactory  type 
for  the  work  referred  to  in  the  preceding  question. 

In  a  "series"  machine  the  current  generated  in  the  armature  passes 
in  equal  strength  through  the  field  magnets  and  external  circuit. 

A  "shunt"  wound  dynamo  differs  from  the  "series"  machine  in  that 
only  a  comparative  small  part  of  the  current  generated  by  the  arma- 
ture is  used  for  the  excitation  of  the  field  coils,  the  greater  part  of  the 
current  being  led  direct  to  the  external  circuit.  By  putting  more  or 
less  resistance  in  the  "fields"  of  these  machines,  the  electro-motive  force 
at  the  terminals  thereof  may  be  varied. 

A  compound  wound  dynamo  combines  in  its  fields  the  winding  of 
both  the  series  and  shunt  machines.  They  may  be  wound  to  keep  a 
constant  electro-motive  force  at  varying  loads  or  may  be  "over-com- 
pounded" so  as  to  allow  for  loss  in  feeders.  This  is  the  modern  genera- 
tor used  for  direct  current  incandescent  lighting  and  for  street  rail- 
way work. 

Q.  65.  (1898-9.)  A  25  HP  motor  is  installed  to  drive  an  isolated  por- 
tion of  a  factory;  30  arc  lamps  at  5  amperes  each,  and  300  16-c.p.  incan- 
descent lamps  are  required  for  lighting.  What  should  be  the  capacity 
of  the  dynamo,  in  kilowatts,  to  furnish  readily  the  current  required 
for  all? 

Ans.  65.  One  horse-power  is  equivalent  to  746  watts,  the  latter  unit 
being  the  one  used  to  express  work  done  by  electrical  power.  The 
kilowatt  is  a  multiple  of  this  unit,  its  value  being  1,000  watts. 

For  the  case  in  hand  the  power  required  for  a  25  HP  motor  is 
25  X  746  =  18,650  watts. 

Thirty  arc  lamps  at  5  amperes  each  =  150  amperes. 

Watts  =  amperes  X  volts;   or  150  X  115  —  17,250  watts. 

One  16-candle  power  incandescent  lamp  =  56  watts,  hence  for  300 
lamps  we  have  300  X  56  =  16,800  watts. 

Summary: 

25  HP  motor  =  18,650  watts. 

30  arc  lamps  =  17,250  watts. 

300   incandescent  =  16,800  watts. 


52,700 

or  52.7  kilowatts. 

Losses  from  various  causes  may  be  estimated  at  15%,  therefore  a 
K.  W.  generator  is  required  for  the  work  indicated  in  the  question. 


Q.  66  (1898-9.)  Give  the  electrical  horse-power  equivalent  to  the 
work  noted  in  Q.  65  and  state  the  engine  horse-power  you  would  pro- 
vide to  drive  same  satisfactorily. 

rSe"POWer  6quivalent  to  work   r^rred  to   in  the 
60,000  watts 


=  80  HP. 

746 

152 


An  engine  developing  100  actual  horse-power  will  give  satisfactory 

results.  

Q.  73.  (1898-9.)  Explain  the  difference  and  advantages  of  bi-polar 
and  multi-polar  arc  dynamos. 

Ans.  73.  We  know  of  no  build  of  multipolar  "arc"  machines.  All 
arc  machines  so  far  constructed  that  we  know  anything  about  are  bi- 
polar or  their  equivalent.  (The  word  "arc"  in  the  question  is  a  pal- 
pable error. — Ed.  Com.) 

Incandescent  machines  are  mostly  multi-polar  and  are  now  made 
that  way  in  the  larger  sizes;  their  advantages  are  simplicity  in  con- 
struction of  frame  and  field  winding,  facility  of  access  in  case  01  needed 
repairs  and  slower  speed  of  armature  revolution. 


Q.  74.  (18989.)  Name  the  proper  material  and  the  style  of  brushes 
best  adapted  for  bi-polar  and  multi-polar  machines;  also  give  various 
causes  for  "sparking." 

Ans.  74.  Bi-polar  machines  were  first  fitted  with  copper  brushes  and 
most  old  machines  of  that  type  are  unsuitable  for  carbon  brushes  on 
account  of  their  smaller  commutator  surface  and  consequent  small 
area  allotted  to  a  brush  of  this  kind,  causing  much  heating  and  dissat- 
isfaction. . 

All  multi-polar  generators  have  liberal  commutators  and  large  brush 
area,  hence  carbon  brushes  are  used  .and  invariably  with  much  satis- 
faction. 

Causes  for  "sparking"  are  varied  and  many;  it  may  be  due  to  the 
following  among  other  causes: 

(a)  Brushes  may  not  be  set  at  point  of  commutation. 

(b)  Brushes  may  be  wedged  in  holders. 

(c)  Brushes  may  not  be  properly  fitted  to  commutator. 

(d)  Brushes  may  not  bear  with  sufficient  pressure. 

(e)  Brushes  may  be  burned  on  ends. 

(f)  Commutator  may  be  rough. 

(g)  Commutator  may  have  a  loose  or  projecting  bar. 
(h)     Commutator  may  be  dirty,  oily  or  worn. 

(i)     Machine  may  be  badly  designed  or  overloaded. 

(j)     Loose  connection  of  coil  to  commutator,  or  open  circuit. 


Q.  75.  (1898-9.)  State  in  a  general  way  how  a  dynamo  should  be 
connected  to  switch-board,  how  circuits  should  be  run  from  same,  and 
where  circuit-breakers  and  fuses  should  be  placed,  explaining  why  the 
latter  are  put  in. 

Ans.  75.  The  exact  arrangement  of  the  switch-board  varies  with  the 
judgment  of  the  designer  or  the  requirements  of  the  case.  Dynamos 
are  generally  connected  from  the  terminal  board  to  the  switch-board  by 
cables.  Generally  the  positive  lead  is  connected  to  one  leg  of  the  switch. 
The  circuit  switches  are  connected  to  the  main  switch  by  means  of 
bus-bars  which  take  the  entire  current  from  the  main  switch  and 
distribute  it  to  the  different  circuit  switches. 

Fuses  and  circuit  breakers  are  used  to  protect  the  lines  and  machines 
from  burning  out,  due  to  excessive  current  caused  by  short  circuits  or 
an  overload,  either  momentary  or  sustained.  They  are  generally  placed 
betwen  the  machine  and  main  switch;  fuses  are  placed  on  all  distribut- 
ing circuits  between  the  line  and  the  switch. 


Q.  76.  (1898-9.)  What  is  an  automatic  circuit-breaker?  Upon  what 
does  its  action  depend,  and  what  advantages  has  it  over  a  fuse? 

Ans.  76.  An  automatic  circuit  breaker  is  an  apparatus  for  opening 
the  circuit  when  an  excessive  amount  of  current  passes  through  it.  A 
common  form  depends  for  its  action  upon  the  principle  that  a  bar  of 

153 


iron  will  tend  to  balance  itself  in  a  solenoid.  In  this  case  the  solenoid 
is  composed  of  three  or  four  turns  of  the  entire  current  conductor;  as 
the  current  increases,  the  soft  iron  bar  raises  until  it  trips  the  spring 
actuated  switch.  Their  advantage  over  a  fuse  is,  that  in  case  of  a 
momentary  overload  they  can  be  quickly  thrown  in  again,  whereas  a 
fuse  takes  time  to  replace  and,  under  certain  conditions,  is  a  dangerous 
operation.  

Q.  77.  (1898-9.)  What  style  "winding"  is  best  adapted  for  a  motor 
which  is  to  run  at  a  constant  speed  and  carry  a  variable  load? 

Ans.  77.  The  shunt  winding  is  almost  universally  used  for  motors 
with  variable  load  and  using  direct  current.  This  type  of  motor  gives 
a  very  nearly  constant  speed  with  a  variable  load,  providing  the  electro- 
motive force  is  kept  constant. 


Q.  78.  (1898-9.)  Should  motors  be  run  on  independent  circuits? 
Why? 

Ans.  78.  Motors  should  be  run  on  independent  circuits,  very  often, 
however,  they  are  cut  in  almost  anywhere.  The  work  of  a  motor  with 
a  variable  load  requires  more  or  less  current  and  if  not  on  a  separate 
circuit,  unless  the  conductors  are  excessively  large,  will  make  the  lights 
unsteady. 

Q.  79.  (1898-9.)  What  is  a  starting-box,  and  why  required  in  con- 
nection with  a  motor.?  Explain  also  the  automatic  cut-out  and  the  rea- 
sons for  attaching  same. 

Ans.  79.  A  starting  box  is  necessary  in  order  to  insert  a  resistance 
in  the  circuit,  to  allow  the  motor  to  build  up  its  counter  B.  M.  F.,  other- 
wise one  or  more  coils  would  burn  out,  as  the  total  current  would  pass 
through  them  before  the  armature  had  a  chance  to  start  revolving. 

An  automatic  cut-out  is  so  arranged  that  should  the  external  circuit 
be  broken,  the  box  will  automatically  cut  in  its  resistance  and  then  open 
the  armature  circuit.  ;  (*' 

-• ;       '•--     «.v.-:»«..:?X*.*3      (ri) 

Q.  80.  (1898-9.)  Name  the  instruments  generally  used  in  connec- 
tion with  an  electrical  plant  and  explain  what  their  readings  indicate. 

Ans.  80.  Ammeter,  or  ampere  meter,  indicates  the  amount  of  cur- 
rent or  amperage.  The  volt-meter  indicates  the  electrical  pressure  on 
the  line,  or  the  voltage.  The  ground  detector,  or  pilot  lamps,  indicate  any 
short  circuits  on  the  line. 

A  watt-meter  should  be  in  every  station  where  it  is  desirable  to 
accurately  determine  the  "out-put"  of  the  plant.  - 


Q.  81.  (1898-9.)  What  general  rules  should  be  obseraed :  in  .making 
electrical  connections?  -./  -j  dfoijdv  -t^ •;-;,/:.• 

Ans.  81.  The  metallic  surfaces  should  be  perfectly  clean  and  make 
good  Contact.  They  should  also  be  mechanically  strong  enough  to 
'withstand  any  strain  likely  to  be  encountered.  .,:.•  ';  . . 


Q.  82.  (1898-9.)  What  broad  rule  should  be  observed  on'the  sctire 
of  personal  safety  when  handling  electrical  conductors? 

Ans.  82.  Be  careful  not  to  interpose  any  portion  of  the  body  where 
such  will  become  a  conductor  of  current. 

A  good  rule  for  personal  safety  when  working  around  eleoteical 
machines  is  to  wear  rubber  gloves;  use  one  hand  only,  as-muoteos 
possible,  and  always  stand  on  a  dry  or  insulated  spot.  •'  /icimnoo 


Q.  1.  (1899-1900.)  Define  electricity.  From  what  is  the  term  de- 
rived? State  its  nature  as  demonstrated  by  recent  research,  and  upon 
what  is  electrical  science  founded? 

Ans.  1.  Electricity  is  the  name  given  to  the  cause  of  all  electrical 
phenomena,  derived  from  the  Greek  word  "electron,"  meaning  amber. 
In  the  German  this  substance  is  called  "bernstein,"  translated  "burn- 
stone."  A  fossil  indurated  vegetable  juice,  transparent  or  translucent, 
sometimes  colorless,  but  usually  some  shade  of  yellow  and  brown;  nega- 
tively electrified  by  friction  when  rubbed  with  fur  or  flannel.  Amber 
being  susceptible  to  a  high  degree  of  polish,  it  probably  was  in  this 
way  that  its  peculiar  properties  of  attraction  and  repulsion  was  first 
noticed  by  the  Greeks,  which  we  now  understand  to  be  due  to  electricity. 
It  is  found  in  alluvial  soils,  beds  of  lignite  and  on  sea  coasts,  especially 
the  Prussian  coast  of  the  Baltic  Sea. 

Although  electrical  science  has  advanced  sufficiently  far  to  recognize 
the  fact  that  the  exact  nature  of  electricity  is  unknown,  recent  research 
tends  to  demonstrate  that  all  electrical  phenomena  are  due  to  a  peculiar 
strain  or  stress  of  a  medium  called  ether;  that  when  in  this  condition 
the  ether  possesses  potential  energy  or  capacity  for  doing  work  as  is 
manifested  by  attractions  and  repulsions,  by  chemical  decomposition, 
and  by  luminous,  heating  and  various  effects.  In  all  probability,  elec- 
tricity is  not  a  form  of  matter,  for  it  possesses  only  two  physical  prop- 
erties in  common  with  material  substance,  namely  "indestructibility 
and  elasticity;"  it  possesses  neither  weight,  extension  nor  any  of  the 
other  physical  properties  of  matter. 


Q.  2.  (1899-1900.)  What  are  the  two  most  essential  features  in 
acquiring  a  knowledge  of  electrical  science? 

Ans.  2.  -In  acquiring  a  knowledge  of  the  electrical  science,  the  two 
most  essential  features  are,  first,  to  learn  how  to  develop  electrical 
action,  and,  second,  to  determine  the  effects  produced  by  it. 


Q.  3.  (1899-1900.)  The  number  of  processes  for  developing  electrical 
action  is  almost  innumerable,  but  can  be  classified  under  four  heads. 
Name  them. 

Ans.   3.      (1)     By  the  contact  of  dissimilar  substances. 

(2)  By  chemical  action. 

(3)  By  the  application  of  heat. 

(4)  By  magnetic  induction. 


Q.  4.     (1899-1900.)     Name  five  different  principal  ways  in  which  the 
presence  of  electricity  can  be  detected. 

Ans.  4.     (1)     Cause  attractions  and  repulsions  of  light  particles  of 
matter,  such  as  feathers,  pith,  gold  leaf,  pieces  of  paper,  etc. 

(2)  Decompose  certain  forms  of  matter  into  their  various  elements 
and  cause  other  chemical  changes. 

(3)  Produce  motion  in  a  freely  suspended  magnetic  needle,  such 
as  the  needle  of  a  compass. 

(4)  Violently  agitate  the  nervous  systems  of  all  animals. 

(5)  Heat  the  substance  through  which  it  acts. 


Q.  5.  (1899-1900.)  When  electricity  appears  to  reside  upon  the  sur- 
face of  bodies  under  high  tension,  what  branch  of  the  science  treats  of 
this  condition,  and  what  are  the  charges  termed?  When  they  flow 
through  the  substance  of  a  body  under  a  comparatively  low  tension, 
what  branch  of  the  science  treats  of  this  condition? 

Ans.  5.  When  electricity  appears  to  reside  upon  the  surface  under  a 
high  tension,  this  branch  of  the  science  is  termed  "electro-statics,"  and 
the  charges  are  said  to  be  static  charges  of  electricity.  When  the  cur- 
rent appears  to  flow  through  their  substance  under  comparatively  low 
pressure  or  tension,  this  branch  of  the  science  is  termed  "electro-dy- 
namics," and  treats  of  the  action  of  electric  currents. 

155 


Q.  6.  (1899-1900.)  Describe  and  give  some  examples  of  static 
charges.  What  do  you  understand  by  the  term  positive  ( + )  and  nega- 
tive ( )  charges?  Can  a  positive  or  negative  charge  be  produced  inde- 
pendent of  one  another?  Why?  What  about  the  intensity  of  the 
charge  ? 

Ans.  6.  If  a  glass  rod  or  a  piece  of  amber  is  rubbed  with  fur  or  silk 
the  parts  rubbed  have  the  property  of  attracting  light  particles  of  mat- 
ter. These  attractions  and  repulsions  are  due  to  a  static  charge  residing 
upon  the  surface  of  these  bodies,  and  a  body  in  this  condition  is  said 
to  be  electrified. 

An  apparatus  known  as  an  electric  pendulum  is  often  used  to  demon- 
strate a  number  of  interesting  experiments,  consisting  of  a  pith  ball 
suspended  by  a  silk  string  and  a  glass  rod  or  a  stick  of  sealing-wax, 
electrified  by  different  substances,  exemplifying  the  attractions  and 
repulsions,  according  to  a  positive  and  negative  electrifying  of  the 
substance. 

An  electric  charge  developed  upon  glass  by  rubbing  with  silk  has 
been  termed  a  positive  (  +  )  charge;  and  that  developed  on  resinous 
bodies  rubbed  with  fur  or  flannel,  a  negative  ( — )  charge. 

Neither  a  positive  nor  negative  charge  can  be  produced  independent 
of  one  another,  for  there  is  always  an  equal  quantity  of  both  charges 
produced,  one  charge  appearing  on  the  surface  of  the  body  rubbed  and 
an  equal  amount  of  the  opposite  charge  upon  the  rubber.  The  intensity 
of  the  charge  developed  is  independent  of  the  actual  amount  of  friction 
which  takes  place  between  the  two  bodies. 

To  obtain  the  nighest  possible  degree  of  electrification  it  is  only 
necessary  to  bring  every  portion  of  one  surface  into  contact  with  every 
particle  or  portion  of  the  other;  when  this  is  done  no  extra  amount 
of  rubbing  can  develop  any  greater  charge  upon  either  surface. 


Q.  7.  (1899-1900.)  From  experiments  with  static  charges  give  two 
laws  governing  the  phenomena. 

Ans.  7.  (1)  When  two  dissimilar  substances  are  brought  into 
contact,  one  of  them  always  assumes  the  positive  and  the  other  the  neg- 
ative condition,  although  the  amount  may  sometimes  be  so  small  as  to 
render  its  detection  difficult. 

(2)  Electrified  bodies  with  similar  charges  are  mutually  repellant, 
while  electrified  bodies  with  dissimilar  charges  are  mutually  attractive. 


Q.  8.  (1899-1900.)  Name  a  few  of  the  list  known  as  the  electric 
series,  and  the  relation  they  bear  to  each  other. 

Ans.  8.  The  following  is  the  list  of  the  electric  series,  and  the  sub- 
stances so  arranged  that  each  receives  a  positive  charge  when  rubbed 
or  placed  in  contact  with  any  of  the  bodies  following  it.  And  a  nega- 
tive charge  when  rubbed  with  any  that  precede  it: 

:,  fur;  2,  flannel;  3,  ivory;  4,  crystals;  5,  glass;  6,  cotton;  7,  silk; 
»,  the  body;  9,  wood;  10,  metals;  11,  sealing-wax;  12,  resin;  13,  sulphur; 
14,  gutta  percha;  15,  gun  cotton. 

For  example,  glass  when  rubbed  with  flannel  receives  a  negative 
charge,  and  when  rubbed  with  sealing-wax  a  positive  charge. 


Q.  9.  (1899-1900.)  Explain  the  terms  conductors,  non-conductors 
nsulators.  Give  a  list  of  good  conductors;  also  of  non-conductops. 

Ans.  9.  Experiments  show  that  when  a  metal  receives  a  charge  at 
any  point,  the  electricity  immediately  passes  or  flows  through  its  sub- 
o^electricit  Pai>tS'  Metals>  therefore.  are  said  to  be  good  conductors 

Experiments  have  also  shown  that  in  the  case  of  the  dry  glass  rod 

«nhT       a  ^-e(lev0f  EP**1?**  or  resin'  that  onlv  that  Part  of  the 
substance  which  has  been  rubbed  will  be  electrified,  the  other  parts  will 

156 


produce  neither  attractions  nor  repulsions.  These  bodies  do  not  readily 
conduct  electricity;  that  is,  they  oppose  or  resist  the  passage  of  elec- 
tricity through  them  and  are  termed  non-conductors  of  electricity. 

This  distinction  is  not  absolute,  for  all  bodies  to  some  extent  conduct 
electricity,  while  there  is  no  known  substance  but  what  offers  some 
resistance  to  the  flow  of  the  electric  current. 

The  names  of  some  of  the  most  important  in  the  list  of  good  conduc- 
tors and  non-conductors  are  given,  classified  as  to  their  conductivity, 
etc.: 

Conductors — Silver,  copper,  aluminum,  other  metals,  charcoal,  water, 
the  body. 

Non-conductors — Paper,  oils,  porcelain,  wood,  silk,  resins,  gutta 
percha,  shellac,  ebonite,  parafflne,  glass,  dry  air,  etc. 


Q.  10.  (1899-1900.)  Define  the  word  potential.  For  what  is  it  sub- 
stituted and  to  what  is  it  analogous? 

Ans.  10.  Potential  is  a  word  substituted  for  the  general  and  vague 
phrase,  electrical  condition.  It  is  analogous  with  pressure  in  gases; 
heads  in  liquids,  and  temperature  in  heat. 


Q.  11.  (1899-1900.)  Explain  the  conditions  when  an  electrified  body 
is  connected  to  earth — direction  of  flow  as  regards  a  high  or  low  poten- 
tial in  relation  to  earth. 

Ans.  11.  If  an  electrified  body,  positively  charged,  is  connected  to 
the  earth  by  a  conductor,  electricity  is  said  to  flow  from  the  body  to  the 
earth;  but  if  negatively  charged  and  connected  to  the  earth  in  a  similar 
manner,  electricity  is  said  to  flow  from  the  earth  to  that  body.  This 
is  called  the  direction  of  flow  of  an  electric  current. 

That  which  determines  the  direction  of  the  flow  is  the  relative 
electrical  potential,  or  the  pressure  of  the  two  charges  in  regard  >to  the 
earth. 

It  is  impossible  to  say  with  certainty  in  which  direction  electricity 
really  flows,  or,  in  other  words,  to  declare  which  of  two  points  has  the 
higher  or  lower  electrical  potential,  or  pressure. 

All  that  can  be  said  with  certainty  is  that  when  there  is  a  difference 
of  electrical  potential,  or  pressure,  electricity  tends  to  flow  from  the 
higher  to  that  of  the  lower  potential  or  pressure. 

For  convenience  it  has  been  arbitrarily  assumed  and  conventionally 
adopted  that  the  positive  electrical  condition  is  at  a  higher  potential, 
or  pressure,  than  that  called  the  negative,  and  that  electricity  tends  to 
flow  from  a  positively  to  a  negatively  electrified  body. 


Q.  12.     (1899-1900.)     What  is  meant  by  zero  potential? 

Ans.  12.  The  normal  level  of  the  water  is  taken  as  that  of  the  sur- 
face of  the  sea,  the  normal  pressure  of  air  and  gases,  as  that  of  the 
atmosphere  at  the  sea  level;  similarly  there  is  a  zero  potential,  or 
pressure,  of  electricity  in  the  earth  itself. 

The  earth  may  be  regarded  as  a  reservoir  of  electricity  '^r  infinite 
quantity,  and  its  potential,  or  pressure,  taken  as  zero. 

A  positive  electrical  condition  may  be  assumed  to  be  at  a  higher 
potential,  or  pressure,  than  the  earth,  and  that  called  negative  is  as- 
sumed to  be  at  a  lower  potential  than  the  earth. 


Q.  13.  (1899-1900.)  What  is  first  necessary  to  do  in  order  to  pro- 
duce an  electric  current?  What  is  the  effect  of  placing  a  piece  of  cop- 
per and  zinc  together;  when  separated  slightly  and  one  end  of  each 
submerged  in  saline  or  acidulated  water?  Describe  the  simple  voltaic 
or  galvanic  cell. 

Ans.  13.  To  produce  an  electric  current,  it  is  first  necessary  to 
cause  a  difference  of  electrical  potential  between  two  bodies:,  or  be- 
tween two  parts  of  the  same  body. 

157 


When  two  dissimilar  substances  are  placed  in  contact,  one  always 
assumes  the  positive  and  the  other  the  negative  condition. 

There  is  instantly  developed  a  difference  of  potential  between  the 
two  bodies. 

A  piece  of  copper  and  zinc  placed  in  contact  will  develop  a  differ- 
ence of  electrical  potential  easily  detected. 

Should  the  plates  be  slightly  separated,  and  one  end  of  each  sub- 
merged in  a  vessel  containing  saline  or  acidulated  water,  the  same 
results  will  follow.  The  exposed  ends  of  the  zinc  and  copper  are  now 
electrified  to  different  degrees;  that  is,  there  is  a  difference  of  poten- 
tial between  them. 

The  voltaic  cell  is  an  apparatus  for  developing  a  continuous  current 
of  electricity  and  consists  essentially  of  a  vessel  containing  saline  or 
acidulated  water,  into  which  are  submerged  two  plates  of  dissimilar 
metals,  or  one  metal  and  a  metalloid. 

The  exposed  ends  are  connected  by  any  conducting  material,  the 
potential  between  the  plates  tends  to  equalize  and  a  momentary  dis- 
charge passes  between  the  exposed  ends  through  the  conductor  and  also 
between  the  submerged  ends  through  the  liquid. 

During  the  passage  of  the  current  through  the  liquid,  it  causes  cer- 
tain chemical  changes  to  take  place;  these  changes  in  their  turn  show 
a  fresh  difference-  of  potential  between  the  plates,  followed  instantly 
by  another  equalizing  discharge. 

These  changes  follow  one  another  so  rapidly  that  they  form  an 
almost  continuous  current. 

.An  electric  current  becomes  continuous  when  the  difference  of  the 
potential  is  constantly  maintained. 

The  name  given  to  the  liquid  in  which  the  metals  are  submerged  is 
"electrolyte,"  which,  as  it  transmits  the  current,  is  decomposed  by  it. 

Of  the  submerged  ends,  the  zinc  assumes  the  positive  condition  and 
the  copper  the  negative;  of  the  exposed  ends  the  copper  is  the  positive 
and  the  zinc  the  negative;  hence  the  flow  of  the  current  is  from  the 
submerged  end  of  the  zinc  through  the  electrolyte  to  the  submerged 
end  of  the  copper,  thence  from  the  exposed  end  of  the  copper  through 
a  conductor  forming  any  outside  circuit  to  the  exposed  end  of  the  zinc, 
flowing  continuously  when  the  circuit  is  closed,  or  remaining  passive 
when  the  circuit  is  open. 

Q.  14.  (1899-1900.)  In  any  voltaic  couple  which  is  the  positive  and 
which  is  the  negative  element? 

Ans.  14.  Two  dissimilar  metals,  when  spoken  of  separately,  are 
called  voltaic  or  galvanic  elements;  when  taken  collectively,  are  known 
as  a  voltaic  couple.  The  metal  terminals  or  poles  of  a  cell,  to  which 
the  conductors  are  attached,  are  termed  electrodes,  the  polarity  of 
which  are  directly  opposite  to  the  polarity  of  the  submerged  ends  of 
the  metal  substance. 

The  element  which  is  acted  upon  by  the  electrolyte  will  always 
be  the  positive  element,  and  its  electrode  or  terminal  the  negative  elec- 
trode of  the  cell  in  any  voltaic  couple. 


Q.  15.  (1899-1900.)  Electromotive  series;  name  them  in  the  regu- 
lar order  and  tell  how  the  positive  or  negative  element  may  be  deter- 
mined. Give  the  difference  of  potential  in  relation  of  one  element  to 
another  of  the  series. 

Ans.  15.  The  following  list  of  voltaic  elements  composes  what  is 
known  as  the  electromotive  series;  any  two  of  which  form  a  voltaic 
couple  when  submerged  in  an  electrolyte,  the  one  standing  first  on 
the  list  being  the  positive  element  to  the  one  following  it.  And  the 
ifference  of  potential  will  be  the  greater  in  proportion  to  the  distance 
between  the  two  substances  in  the  list. 

For  example,  the  difference  of  potential  developed  between  zinc  and 

158 


platinum  is  greater  than  between  zinc  and  nickel,  or  equal  to  the  sum 
of  the  difference  of  potential  between  zinc  and  nickel  and  between 
nickel  and  platinum. 

The  electromotive  series: 

1,  zinc;  2,  cadmium;  3,  tin;  4,  lead;  5,  iron;  6,  nickel;  7,  bismuth; 
8,  antimony;  9,  copper;  10,  silver;  11,  gold;  12,  platinum;  13,  graphite. 


Q.  16.  (1899-1900.)  Name  three  ways  in  which  electricity  flowing  as 
a  current  differs  from  static  charges. 

Ans.  16.  1.  The  potential  is  much  lower.  2.  Its  actual  quantity 
is  much  greater.  3.  It  is  continuous. 

The  potential  of  a  current  of  electricity  is  comparatively  so  small 
that  a  voltaic  battery  composed  of  a  large  number  of  cells  is  not  suf- 
ficient to  produce  a  spark  of  more  than  two-hundredths  of  an  inch  in 
air;  whereas,  a  rapidly  moving  leather  belt  will  sometimes  produce 
sparks  of  one  to  three  inches. 

If,  however,  the  actual  quantity  of  electricity  is  measured  by  its 
effect  in  decomposing  water,  then  the  quantity  produced  by  a  small 
voltaic  cell  would  give  greater  results  than  that  from  a  large  rapidly 
moving  belt  producing  static  charges  several  inches  in  length. 


Q.  17.  (1899-1900.)  Can  an  electric  current  be  developed  between 
two  non-conducting  substances  as  in  the  case  of  static  charges?  Will 
it  flow  if  the  conductor  is  not  made  entirely  of  conducting  material? 

Ans.  17.  An  electric  current  cannot  be  developed  between  two  non- 
conducting substances,  as  in  the  case  of  static  charges,  and  it  will 
never  flow  unless  the  conducting  path  is  made  entirely  of  conducting 
material. 


Q.  18.  (1899-1900.)  How  many  ways  are  there  of  grouping  the  cells 
of  a  voltaic  battery?  Describe  each,  and  the  effect  upon  the  potential 
for  each  grouping. 

Ans.  18.  There  are  three  different  ways  or  methods  of  connecting 
or  grouping  the  cells  of  a  voltaic  battery:  In  series;  in  parallel  or  mul- 
tiple-arc; in  multiple  series. 

Cells  are  connected  in  series  when  the  positive  electrode  of  the 
first  cell  is  connected  to  the  negative  electrode  of  the  second,  and  the 
positive  electrode  of  the  second  is  connected  to  the  negative  electrode 
of  the  third  cell,  and  so  on,  as  shown  in  the  diagram  Fig.  1. 

This  method  is  used  when  there  is  available  a  large  number  of  low 
potential  cells  and  a  high  potential  is  desired,  usually  used  for  bell 
service  and  long  telegraph  lines  or  other  high  resistance  circuits. 

The  potential  is  increased  on  each  addition  of  cell,  that  is,  if  the 


mm* 


Fig.  1. 

voltage  of  each  cell  is  2  volts;  then  a  grouping  of  12  cells  in  this 
manner  would  develop  24  volts,  other  groupings  in  the  same  ratio. 

Should  we  wish  to  produce  a  strong  current  at  a  low  voltage  or 
potential,  that  is,  when  the  external  resistance  is  low,  as  in  electro- 
plating, the  cells  may  be  connected  in  parallel  or  multiple-arc;  in  this 
method  only  a  part  of  the  total  current  flowing  in  the  main  conductors 
will  pass  through  each  cell. 

The  diagram  Fig.  2  will  show  the  method  of  connection: 

159 


Fig.  2. 

In  a  grouping  of  this  kind,  the  potential  would  be  equal  to  the 
potential  of  a  single  cell,  no  matter  how  many  cells  there  are  in  the 

Should  a  stronger  current  and  a  higher  potential  be  needed,  then 
the  grouping  can  be  in  what  is  termed  multiple-series,  each  group 
being  composed  of  two  or  more  cells  as  the  case  may  be,  connected 
in  series  and  then  connecting  all  the  groups  in  parallel  or  multiple-arc. 

The  diagram  will  readily  show  the  grouping  and  the  effect  on  the 
potential,  etc.,  each  grouping  according  to  the  number  of  cells  in  the 
grouping  determining  the  potential  of  the  current  developed,  in  this 
case  the  voltage  equals  that  of  two  cells  or  four  volts. 


Fig.  3. 


Q.  19.  (1899-1900.)  Describe  a  circuit — broken  or  open,  closed, 
grounded,  external  and  internal. 

Ans.  19.  A  circuit  is  a  path  or  a  conductor,  or  several  conductors 
joined  together,  through  which  an  electric  current  flows  from  a  given 
point  around  the  conducting  path  back  again  to  its  starting  point. 

A  circuit  is  said  to  be  open,  or  broken,  when  its  conducting  ele- 
ments are  discontinued  in  such  a  manner  as  to  prevent  the  current 
from  flowing. 

A  circuit  is  closed  or  completed  when  its  conducting  elements  are 
uninterrupted  or  continuous. 

A  circuit  is  said  to  be  grounded  when  the  earth  forms  a  part  of  its 
conducting  path. 

An  external  circuit  is  that  part  of  a  circuit  which  is  outside  or 
external  to  its  electric  source. 

An  internal  circuit  is  that  part  of  the  circuit  which  is  included 
within  the  electric  source. 

In  a  voltaic  ce^  the  internal  circuit  consists  of  the  two  metallic 
plates,  or  elements,  and  the  electrolyte. 

The  external  circuit,  a  conductor  connecting  the  free  ends  of  the 
electrodes. 


Q.  20.  (1899-1900.)  When  are  conductors  said  to  be  in  series? 
What  do  you  understand  by  a  shunt  or  derived  circuit,  in  parallel  or 
multiple  arc?  With  your  explanation,  furnish  sketch. 

Ans.  20.  Conductors  are  said  to  be  a  series  when  they  are  so 
joined  together  as  to  allow  the  current  to  pass  consecutively  through 
each. 

In  the  diagram  conductors  a,  b,  c  and  d  are  in  series. 


A  shunt  or  derived  circuit  is  one  which  is  divided  into  two  or  more 
branches,  each  transmitting  a  part  of  the  main  current. 

The  separate  branches  are  said  to  be  in  parallel,  or  multiple-arc: 
C 


d 

The  main  current  flowing  through  the  conductor  (a),  dividing  and 
passing  through  the  branches  (b  and  c),  uniting  and  completing  the 
circuit  through  the  conductor  (d). 

The  two  branches  (c  and  b)  are  said  to  be  in  parallel  or  multiple- 
arc. 

Q.  21.  (1899-1900.)  What  are  magnets,  and  to  what  is'  the  term 
magnetism  applied?  Does  magnetism  exist  in  a  natural  state,  and  by 
what  name  is  it  known  in  chemistry?  Where  and  by  whom  was  the 
ore  first  found?  From  what  fact  was  the  name  lodestone  given  to  the 
ore?  What  are  artificial  magnets?  What  magnets  are  called  perma- 
nent magnets?  Describe  the  common  form  of  artificial  or  permanent 
magnets?  Why  is  a  keeper  used? 

Ans.  21.  Magnets  are  substances  which  have  the  property  of  at- 
tracting pieces  of  iron  or  steel,  and  the  term  magnetism  is  applied 
to  the  cause  of  this  attraction. 

Magnetism  exists  in  a  natural  state  in  an  ore  of  iron,  which  is  known 
in  chemistry  as  magnetic  oxide  of  iron,  or  magnetite.  This  magnetic 
ore  was  first  found  by  the  ancients  in  Magnesia,  a  city  in  Asia  Minor; 
hence  substances  possessing  this  property  have  been  called  magnets. 

It  was  also  discovered  that  when  a  small  bar  of  the  ore  is  suspended 
in  a  horizontal  position  by  a  thread  it  has  the  property  of  pointing 
in  a  north  and  south  direction. 

From  this  fact  the  name  lodestone — leading-stone-was  given  to  the 
ore. 

When  a  bar  or  needle  of  hardened  steel  is  rubbed  with  a  piece  of 
lodestone,  it  acquires  magnetic  properties  similar  to  those  of  the  lode- 
stone,  without  the  latter  losing  any  of  its  own  force.  Such  bars  are 
called  artificial  magnets. 

Artificial  magnets  which  retain  their  magnetism  for  a  long  time 
are  called  permanent  magnets. 

The  common  form  of  artificial  or  permanent  magnets  is  a  bar  of 
steel  bent  in  the  shape  of  a  horseshoe  and  then  hardened  and  magnet- 
ized. 

A  piece  of  soft  iron  called  an  armature,  or  a  keeper,  is  placed  across 
the  two  free  ends,  which  helps  prevent  the  steel  from  losing  its  mag- 
netism. 

Q.  22.  (1899-1900.)  What  is  the  result  of  dipping  a  magnet  into 
iron  filings?  Explain  the  neutral  line,  the  poles  and  the  axis  of  mag- 
netism? Explain  the  polarity  of  a  magnet,  the  attraction  and  repul- 
sions in  general. 

Ans.  22.  If  a  bar  magnet  is  dipped  into  iron  filings,  the  filings 
are  attracted  towards  the  two  ends,  and  adhere  to  these  in  tufts;  while 
towards  the  center  of  the  bar,  half  way  between  the  two  ends,  there  is 
no  such  tendency. 

That  part  of  the  magnet  where  there  is  no  apparent  attraction  is 
called  the  neutral  line;  and  the  parts  around  the  two  ends,  where  the 
attraction  is  the  greatest,  are  called  poles.  An  imaginary  line  drawn 
through  the  center  of  the  magnet,  from  end  to  end,  connecting  the  two 
poles  together,  is  the  axis  of  magnetism. 

161 


Using  a  compass,  which  consists  of  a  magnetized  steel  needle,  rest- 
ing upon  a  fine  point,  so  as  to  turn  freely  in  a  horizontal  plane,  the 
needle,  when  not  in  the  vicinity  of  other  magnets,  or  magnetized  iron, 
will  always  come  to  rest  with  one  end  pointing  to  the  north  and  the 
other  end  towards  the  south.  The  end  pointing  northward  is  the  north 
pole,  and  the  opposite  end  is  the  south  pole.  This  polarity  applies  as 
well  to  all  magnets. 

The  attractions  and  repulsions  are  governed  by  the  same  rule  as 
given  in  answer  to  one  of  the  first  questions.  Electrified  bodies  with 
similar  charges  are  mutually  repellant,  while  those  with  dissimilar 
charges  are  mutually  attractive;  so  it  is  with  the  polarity  of  magnets; 
unlike  poles  having  attraction  between  them  and  like  poles  repelling 
one  another.  

Q.  23.  (1899-1900.)  How  is  the  earth  considered  in  relation  to 
magnetism?  Is  it  possible  to  produce  a  magnet  of  one  pole? 

Ans.  23.  The  earth  is  a  great  magnet  whose  magnetic  poles  coin- 
cide nearly,  but  not  quite,  with  the  true  geographical  north  and  south 
poles. 

It  is  impossible  to  produce  a  magnet  with  only  one  pole.  If  a  long 
bar  magnet  is  cut  into  a  number  of  pieces  each  part  will  still  be  a 
magnet  and  have  a  north  and  south  pole. 

Q.  24.  (1899-1900.)  What  have  you  to  say  in  relation  to  magnetic 
substance?  What  substances  are  magnetic  and  what  are  all  other  sub- 
stances called? 

Ans.  24.  Substances  which,  although  not  in  themselves  magnets, 
not  possessing  polar  and  neutral  lines,  are,  nevertheless,  capable  of 
being  attracted  by  a  magnet.  In  addition  to  iron  and  its  alloys,  the 
following  elements  are  magnetic  substances:  nickel,  cobalt,  mangan- 
ese, oxygen,  cerium  and  chromium.  These,  however,  possess  magnetic 
properties  in  a  very  inferior  degree  compared  with  iron  and  its  alloys. 

All  other  known  substances  are  called  non-magnetic  substances. 


Q.  25.  (1899-1900.)  What  is  the  space  surrounding  a  magnet 
called?  The  direction  of  magnetic  attractions  and  repulsions?  How 
may  the  positions  of  lines  of  force  be  determined?  How  may  their 
direction  be  determined? 

Ans.  25.  The  space  surrounding  a  magnet,  in  which  any  magnetic 
substance  will  be  attracted  or  repelled,  is  called  its  magnetic  field,  or 
simply  its  field. 

Magnetic  attractions  and  repulsions  are  assumed  to  act  in  a  defin- 
ite direction,  and  along  imaginary  lines,  called  lines  of  magnetic  force, 
and  every  magnetic  field  is  assumed  to  be  traversed  by  such  lines  of 
force—  in  fact,  to  exist  by  virtue  of  them.  Their  position  may  be  deter- 
mined (in  any  plane)  by  placing  a  sheet  of  paper  over  a  magnet  and 
sprinkling  fine  iron  filings  over  the  paper.  In  the  case  of  a  bar  magnet 
lying  on  its  side,  the  iron  filings  will  arrange  themselves  in  curved 
lines,  extending  from  the  north  to  the  south  poles.  The  nearer  to  the 
poles  the  more  dense  will  be  these  lines  of  force,  consequently  more  of 
the  filings  will  converge  around  them. 

The  magnetic  field  looking  toward  either  pole  of  a  bar  magnet 
would  exhibit  merely  radial  lines,  and  as  in  the  case  of  the  bar  mag- 
net, the  lines  of  force  will  be  more  dense  at  the  immediate  poles. 

Every  line  of  force  is  assumed  to  pass  out  from  the  north  pole,  mak- 
ing a  complete  circuit  through  the  surrounding  medium  and  into  the 
south  pole  thence  through  the  magnet  to  the  north  pole  again.  This 
is  called  the  direction  of  the  lines  of  force,  and  the  path  which  they 
take  is  called  the  magnetic  circuit. 

By  the  use  of  a  compass  the  direction  of  the  lines  of  force  can  be 
traced;  the  north  pole  of  the  needle  will  always  point  in  the  direction 


162 


Q.  26.  (1899-1900.)  Can  lines  of  force  intersect  each  other?  What 
would  be  the  result  of  two  north  poles,  or  two  south  poles,  being 
brought  near  one  another;  or,  in  other  words,  two  opposing  magnetic 
fields?  What  are  the  resulting  poles  called? 

Ans.  26.  The  lines  of  force  can  never  intersect  each  other;  when 
two  opposing  magnetic  fields  are  brought  together  the  lines  of  force 
from  each  will  be  crowded  and  distorted  from  their  original  direction 
until  they  coincide  in  direction  with  those  opposing,  and  form  a  result- 
ant field  in  which  the  direction  of  the  lines  of  force  will  depend  upon 
the  relative  strength  of  the  two  opposing  negative  fields,  the  resulting 
poles  thus  formed  are  called  consequent  poles.  This  would  be  the  re- 
sult of  two  north  or  two  south  poles  being  brought  near  each  other. 


Q.  27.  (1899-1900.)  What  does  every  magnetic  field  possess?  When 
a  magnetic  substance  is  brought  into  a  magnetic  field,  what  is;  the 
tendency  of  the  lines  of  force?  If  the  substance  is  free  to  move  on  an 
axis,  but  not  bodily  towards  the  magnetic  pole,  in  what  direction  will  it 
come  to  rest?  What  will  the  body  then  become?  What  position  will 
the  south  pole  assume? 

Ans.  27.  In  every  magnetic  field  there  are  certain  stresses  which 
produce  a  tension  along  the  lines  of  force  and  a  pressure  across  them; 
that  is,  they  tend  to  shorten  themselves  from  end  to  end,  and  repel  one 
another  as  they  lie  side  by  side. 

When  a  magnetic  substance  is  brought  into  a  magnetic  field  the 
lines  of  force  in  that  vicinity  crowd  together  and  all  tend  to  pass 
through  the  substance. 

If  the  substance  is  free  to  move  on  an  axis  (but  not  bodily)  towards 
the  magnetic  pole,  it  will  always  come  to  rest  with  its  greatest  extent 
or  length  in  direction  of  the  lines  of  force.  The  body  will  then  become 
a  magnet,  its  south  pole  being  situated  where  the  lines  of  force  enter  it, 
and  its  north  pole  where  they  pass  out. 


Q.  28.  (1899-1900.)  What  is  understood  by  magnetic  induction? 
How  is  the  quantity  of  magnetism  expressed?  What  is  meant  by  the 
term  magnetic  density? 

Ans.  28.  The  production  of  magnetism  in  a  magnetic  substance  in 
this  manner  is  called  magnetic  induction.  The  production  of  artificial 
magnetism  in  a  hardened  steel  needle  or  bar  by  contact  with  lodestone 
is  one  case  of  magnetic  induction.  The  amount  or  quantity  of  mag- 
netism is  expressed  by  the  total  number  of  lines  of  force  contained  in 
the  magnetic  circuit. 

Magnetic  density  is  the  number  of  lines  of  force  passing  through  a 
unit  area  measured  perpendicularly  to  their  direction. 


Q.  29.  (1899-1900.)  Explain  how  the  direction  of  a  current  can  be 
determined  by  the  use  of  a  compass?  If  a  conductor  conveying  a  cur- 
rent should  be  brought  up  through  a  piece  of  card-board  and  iron  filings 
are  sprinkled  on  the  cardboard,  how  will  the  filings  arrange  them- 
selves? What  can  every  conductor  conveying  a  current  be  imagined 
to  be  surrounded  with?  How,  or  in  what  manner,  does  the  magnetic 
density  increase  or  decrease?  Has  the  subject  of  this  question  any- 
thing to  do  with  the  insurance  rules,  in  regard  to  the  separation  of 
conductors,  or  in  other  words,  the  distance  conductors  shall  be  placed 
from  each  other? 

Ans.  29.  If  a  conductor  be  placed  parallel  to  the  magnetic  axis  of 
a  compass  needle,  and  a  current  passed  through  it  in  either  direction, 
the  needle  will  tend  to  place  itself  at  right  angles  to  the  conductor,  or, 
in  general,  an  electric  current  and  a  magnet  exert  natural  force  upon 
each  other. 

Should  the  conductor  be  threaded  up  through  a  piece  of  cardboard 
and  iron  filings  are  sprinkled  on  the  cardboard,  they  will  arrange, 
themselves  in  concentric  circles  around  the  conductor, 

163 


•  This  effect  will  be  observed  throughout  the  entire  length  of  the 
conductor,  and  is  caused  entirely  by  the  current.  In  fact  every  con- 
ductor conveying  a  current  of  electricity  can  be  imagined  as  completely 
surrounded  by  a  sort  of  magnetic  whirl,  the  magnetic  density  decreas- 
ing-as  the  distance  from  the  current  increases. 

The  National  Board  of  Fire  Underwriters,  in  their  national  code 
for  the  installation  of  electric  machinery  and  wiring  of  buildings  have 
promulgated  certain  rules  for  the  separation  of  conductors  conveying 
currents  of  electricity.  This  magnetic  influence,  together  with  other 
reasons,  have  most  certainly  entered  into  the  problem  upon  which 
they  have  made  their  decision  in  the  matter.  Granting  that  the  proper 
insulation  of  the  conductor  is  calculated  to  eliminate  all  danger  from 
magnetic  influence,  contacts  are  often  made  through  abrasions  of  the 
insulation  of  the  conductor,  which  often  happens  through  accident, 
carelessness  or  other  means,  the  wire  becoming  detached  from  its  fast- 
Qning,  swaying  or  pushed  out  of  its  proper  position,  etc. 

In  substantiation  of  this  statement  we  add  a  quotation  from  their 
general  suggestions:  "In  all  electric  work  conductors,  however  well 
insulated,  should  always  be  treated  as  bare,  to  the  end  that  under  no 
conditions,  existing  or  likely  to  exist,  can  a  grounding  or  short  circuit 
occur,  and  so  that  all  leakage  from  conductor  to  conductor,  or  between 
conductor  and  ground,  may  be  reduced  to  the  minimum." 


Q.  30.  (1899-1900.)  If  the  current  is  flowing  away  from  the  ob- 
server, in  which  direction  will  the  lines  of  force  encircle  the  conduc- 
tor? 

Ans.  30.  If  the  current  is  flowing  in  a  conductor  away  from  the 
observer,  then  the  direction  of  the  lines  of  force  will  encircle  the  con- 
ductor from  left  to  right,  or  in  the  same  direction  as  the  hands  of  a 
clock  move. 


Q.  31.  (1899-190X).)  Explain  in  either  case  how  two  parallel  con- 
ductors, both  transmitting  currents  of  electricity,  would  be,  either 
mutually  attractive  or  repellant?  Suppose  the  conductor  was  bent 
into  a  loop,  what  directions  would  the  lines  of  force  take? 
',  Ans.  31.  Two  parallel  conductors,  both  transmitting  currents  of 
electricity,  are,  mutually  attractive  or  repellant,  depending  upon  the 
relative  direction  of  their  currents.  If  the  current  in  both  conductors 
ar^e  flowing  the  same  direction  the  lines  of  force  will  tend  to  surround 
Tjoth  conductors  and  contract,  thus  attracting  both  conductors.  If 
however,  the  currents  are  flowing  in  opposite  directions,  the  lines  of 
orce  lying  between  the  conductors  will  have  the  same  direction  and 
Tore,  repel  the  conductors. 

the  conductor  be  bent  into  a  loop  then  all  the  lines  of  force 
conductor  will  thread  through  the  loop  in  the  same  direc- 


s  (18"-1900-)     Describe  the   helix.    How   would   the  lines   of 
.    coincide  with  the  different  loops?    How  in  relation  to  the  helix 
as  a  wnoie :     r  urnish  sketch. 

Ans.  32.    A  helix  is  formed  by  bending  a  conductor  into  a  series 


of  loops,  or  coil,  the  lines  of  force  around  each  loop  will  coincide 
with  those  around  the  adjacent  loops,  forming  several  long  lines  of 
force  which  thread  through  the  entire  helix,  entering  at  one  end  and 
passing  out  at  the  other  end. 

The  same  conditions  existing  in  the  helix  as  exist  in  the  bar  magnet, 
the  lines  of  force  pass  out  from  one  end  and  enter  the  other  end,  in 
fact  having  a  north  and  south  pole,  a  neutral  line,  and  all  the  properties 
of  attraction  and  repulsion  of  a  magnet. 


Q.  33.  (1899-1900.)  If  a  helix  is  suspended  in  a  horizontal  position, 
and  free  to  turn,  in  what  direction  will  it  come  to  rest? 

Ans.  33.  A  helix  suspended  in  a  horizontal  position  and  free  to 
turn,  it  will  come  to  rest  in  a  north  and  south  direction  if  a  current  be 
passing  through  it. 

Q.  34.  (1899-1900.)  Describe  and  furnish  sketch  of  a  solenoid. 
Upon  what  does  the  polarity  of  a  solenoid  depend? 

Ans.  34.  A  helix  made  in  the  manner  described  in  Answer  32  and 
around  which  a  current  of  electricity  is  circulating,  is  called  a  solen- 
oid; that  is,  the  solenoid  is  the  magnetizing  coil  of  an  electro  magnet. 


LJ^L^L^Wa^1~1""  N 


The  polarity  of  a  solenoid,  that  is,  the  direction  of  the  lines  of  force 
which  thread  through  it,  depends  upon  the  direction  in  which  the 
conductor  is  coiled,  and  the  direction  of  the  current  in  the  conductor. 

To  determine  the  polarity  of  a  solenoid,  knowing  the  direction  of 
the  current:  If  in  looking  at  the  end  of  the  helix,  it  is  so  wound  that 
the  current  encircles  the  helix  in  the  direction  of  the  hands  of  a  clock, 
that  end  will  be  the  south  pole;  if  in  the  other  direction  it  will  be 
the  north  pole.  . 

Q.  35.  (1899-1900.)  In  looking  at  the  end  of  a  helix,  if  it  is  so 
wound  that  the  current  circulates  around  the  helix  from  left  to  right, 
which  end  will  be  the  north  pole?  If  wound  so  that  the  current  circu- 
lates around  from  right  to  left,  which  end  will  be  the  south  pole? 
In  either  case,  in  which  direction  will  the  lines  of  force  take  through 
the  helix? 

Ans.  35.  In  looking  at  the  end  of  a  helix,  so  wound  that  the  cur- 
rent encircles:  the  helix  from  left  to  right,  or  clockwise,  the  end  away 
from  the  observer  will  be  the  north  pole.  If  wound  so  that  the  current 
encircle  the  helix  from  right  to  left,  the  end  away  from  the  observer 
will  be  the  south  pole.  In  either  case  the  lines  of  force  will  leave  the 
north  pole  and  pass  around  and  in  at  the  south  pole,  through  the  helix 
to  the  north  pole  again. 


Q.  36.   (1899-1900.)     How  can  the  polarity  of  a  solenoid  be  changed? 
Ans.  36.     The  polarity  of  a  solenoid  can  be  changed  by  reversing 
the  direction  of  the  current  in  the  conductor. 


Q.  37.  (1899-1900.)  Why  does  a  magnetic  substance  offer  a  better 
path  for  the  lines  of  force  than  air  or  other  non-magnetic  substances? 
What  do  you  understand  by  magnetic  permeability? 

165 


Ans  37  A  magnetic  substance  offers  a  better  path  for  the  lines  of 
force  for  the  same  reasons  explained  in  relation  to  magnets  in  offering 
a  better  path  than  air  or  other  non-magnetic  substances. 

The  facility  offered  by  any  magnetic  substance  to  the  passage 
through  it  of  the  lines  of  force  is  called  magnetic  permeability. 

The  permeability  of  all  non-magnetic  substances,  such  as  air,  copper, 
wood,  etc.,  is  taken  as  1  or  unity. 

The  permeability  of  soft  iron  may  be  as  high  as  2,000  times  that  of 
air. 

If  therefore,  a  piece  of  soft  iron  be  inserted  into  the  magnetic  cir- 
cuit of  a  solenoid,  the  number  of  lines  of  force  will  be  greatly  increased, 
and  the  iron  will  become  highly  magnetized. 


Q.  38.  (1899-1900.)  What  is  an  electro  magnet?  What  is  the  sub- 
stance around  which  the  current  circulates  called?  How  is  the  solen- 
oid generally  termed? 

Ans.  38.  A  magnet  produced  by  inserting  a  magnetic  substance 
into  the  magnetic  circuit  of  a  solenoid  is  an  electro-magnet,  and  the 
substance  around  which  the  current  flows  or  circulates  is  called  the 
core.  The  solenoid  is  generally  termed  the  magnetizing  coil. 


Q.  39.  (1899-1900.)  Describe  the  ordinary  form  of  the  electro-mag- 
net; its  winding,  insulation  and  why  insulation  is  necessary. 

Ans.  39.  In  the  ordinary  form  of  electro-magnets  the  magnetizing 
coil  consists  of  a  large  number  of  turns  of  insulated  wire,  that  is,  wire 
covered  with  a  layer  or  coating  of  some  non-conducting  or  insulating 
material,  usually  cotton  or  silk;  otherwise  the  current  would  take  a 
shorter  and  easier  circuit  from  coil  to  coil  througn  the  iron  core  with- 
out circulating  around  it;  for  this  reason  each  coil  is  thoroughly  in- 
sulated from  one  another  that  the  current  may  be  forced  to  make  the 
requisite  number  of  turns  around  the  core  for  such  specific  purposes 
that  the  coil  may  be  designed  for. 


Q.  40.  (1899-1900.)  How  would  the  field  coil  of  a  motor  be  affected 
if  at  some  point  in  the  winding  of  the  coil  there  should  be  a  defective 
place  in  the  insulation,  and  the  current  jumped  through  a  portion  of 
the  winding  instead  of  making  the  regular  number  of  turns?  Would 
it  increase  or  decrease  the  magnetism  of  that  particular  coil?  What 
effect  would  it  have  on  the  remaining  coils  and  how  would  it  affect 
the  speed  of  the  motor  carrying  its  full  rated  load?  How  would  you 
detect  this  particular  trouble?  What  would  be  liable  to  happen  if  the 
motor  should  continue  to  run  until  it  burned  out? 

Ans.  40.  If,  in  a  shunt  wound,  constant  potential  motor,  one  of 
the  field-coils:  should  be  partially,  or  wholly,  short-circuited,  the  mag- 
netism in  that  magnet  would  be,  according  to  the  circumstance,  below 
the  normal,  the  field  current  would  be  excessive,  causing  the  remain- 
ing coils  to  heat  excessively,  the  defective  coil  remaining  comparatively 
cool.  The  counter  B.  M.  P.  would  drop,  the  applied  B.  M.  F.  would 
become  excessive,  and  the  speed  of  the  motor  would  be  increased  to  a 
point  where  the  counter  E.  M.  F.  would  again  balance  the  load. 

This  excessive  applied  E.  M.  F.  would  cause  the  armature  also  to 
heat,  and  should  the  defect  not  be  remedied,  and  the  motor  continue 
to  run  with  a  full  load,  the  consequence  would  be  the  loss  of  one  or 
all  the  good  field  coils,  with,  in  all  probability,  the  armature. 


nJV1!'  (18"-1900-)     Wh*t   are   the  three   principal   units   used   in 
practical  measurements  of  a  current  of  electricity,  and  what  do  they 


denote? 

1 66 


How  would  you  explain  them  in  analogy  with  the  flow  of  fluids  ? 

How  would  they  correspond  with  a  current  flowing  through  a  wire? 

Ans.  41.  The  three  principal  units  used  in  practical  measurements 
of  a  current  of  electricity  are: 

The  ampere,  or  the  practical  unit,  denoting  the  rate  of  flow  of  an 
electric  current,  or  the  strength  of  an  electric  current. 

The  ohm,  or  the  practical  unit  of  resistance. 

The  volt,  or  the  practical  unit  of  electrical  potential  or  pressure. 

Explained  in  analogy  with  the  flow  of  liquids  or  water,  the  force 
which  causes  the  water  to  flow  through  the  pipe  is  due  to  the  head  or 
pressure;  that  which  resists  the  flow,  is  friction  of  the  water  against 
the  inside  of  the  pipe,  and  the  amount  would  vary  with  circumstances. 

The  rate  of  flow,  or  the  current,  may  be  expressed  in  gallons  per 
minute,  and  is  the  ratio  between  the  head,  or  pressure,  and  the  resist- 
ance of  the  water  against  the  inside  of  the  pipe.  For,  as  head  or  pres- 
sure increases,  the  rate  of  flow  increases  in  proportion;  as  the  resist- 
ance increases  the  current  diminishes. 

In  the  case  of  an  electric  current  flowing  through  a  conductor,  the 
electromotive  force,  or  potential,  corresponds  to  the  pressure,  or  the 
head  of  water,  and  the  resistance  which  a  conductor  offers  to  the  flow 
of  electricity  to  the  friction  of  the  water  against  the  pipe. 


Q.  42.  (1899-1900.)  The  strength  of  a  current,  or  the  rate  of  flow 
of  electricity,  is  a  ratio.  Explain  this  ratio? 

By  whom  was  this  ratio  first  discovered?  By  whom  was  it  first 
applied  to  electricity,  and  by  what  name  has  it  since  been  called? 

Ans.  42.  The  strength  of  an  electric  current,  or  the  rate  of  flow 
of  electricity,  is  also  a  ratio — a  ratio  between  the  electromotive  force 
and  the  resistance  of  the  conductor  through  which  the  current  is  flow- 
ing. 

This  ratio  as  applied  to  electricity  was  first  discovered  by  Dr.  G. 
S.  Ohm,  and  has  since  been  called  "Ohm's  Law." 


Q.  43.  (1899-1900.)  What  is  "Ohm's  law"  and  how  is  it  usually 
expressed  algebraically? 

What  is:  an  ampere? 

As  electricity  possesses  neither  weight  nor  extension,  therefore,  it 
cannot  be  measured  like  fluids  or  gases,  how  can  the  strength  of  an 
electric  current  be  determined? 

Ans.  43.  Ohm's  Law. — The  strength  of  an  electric  current  in  any 
circuit  is  directly  proportional  to  the  electromotive  force  developed  in 
that  circuit  and  inversely  proportional  to  the  resistance  of  the  circuit, 
i.  e.,  it  is  equal  to  the  electromotive  force  divided  by  the  resistance. 

Ohm's  Law  is  usually  expressed  algebraically,  thus: 
Electromotive  force 

Strength  of  current  =  

Resistance 

If  C  =  the  current  in  amperes. 

E  =r  the  electromotive  force  in  volts. 

R  =  the  resistance  in  ohms. 

The  formula  will  give  the  strength  of  the  current  (C)  directly  in 
amperes: 

E 

Thus  C  =  — 
R 

An  ampere  is  the  unit  of  strength  of  the  current. 

The  strength  of  an  electric  current  can  be  described  as  a  quantity 
of  electricity  flowing  continuously  every  second,  or,  in  other  words,  it 
is  the  rate  of  flow  of  electricity,  just  as  the  current  expressed  in  gal- 
lons per  minute  is  the  rate  of  flow  of  liquids. 

167 


When  one  unit  quantity  of  electricity  is  flowing  continuously  every 
second  then  the  rate  of  flow,  or  the  strength  of  current,  is  one  ampere; 
if  two  units  quantities  of  current  are  flowing  continuously  every  sec- 
ond then  the  rate  of  flow,  or  the  strength  of  current,  is  two  amperes, 

m  ItS<makes  no  difference  in  the  number  of  amperes  whether  the  cur- 
rent flows  for  a  long  time  or  for  only  a  fraction  of  a  second;  if  the 
quantity  of  electricity  that  would  flow  in  one  second  is  the  same  in 
both  cases,  then  the  current  strength  in  amperes  is  the  same. 

Electricity  possesses  neither  weight  nor  extension,  therefore  an 
electric  current  can  not  be  measured  by  the  usual  method  adopted  for 
measuring  liquids  and  gases. 

In  liquids  the  strength  of  the  current  is  determined  by  measuring 
or  weighing  the  actual  quantity  of  the  liquid  which  has  passed  between 
two  points  in  a  certain  time  and  dividing  the  results  by  that  time. 

Thus,  if  100  gallons  of  water  should  pass  through  a  pipe  in  5  sec., 
the  rate  of  flow  would  be  equal  to  100  H-  5,  or  20  gals,  per  second. 

The  strength  of  an  electric  currant,  on  the  contrary,  is  determined 
directly  by  the  effect  it  produces,  and  the  actual  quantity  of  electricity 
which  has  passed  between  two  points  in  a  certain  time  is  afterwards 
calculated  by  multiplying  the  strecgth  of  the  current  by  the  time. 
The  principal  effects  of  an  electric  current  were  given  in  answer  to  the 
first  set  of  these  questions.  Of  these,  the  one  most  generally  used  for 
measuring,  is  the  action  of  the  current  upon  a  magnetic  needle. 


Q.  44.  (1899-1900.)  Give  a  brief  description  and  the  use  of  a  gal- 
vanometer. 

Ans.  44.  The  instrument  commonly  used  in  laboratory  practice  is 
called  a  galvanometer. 

The  action  of  the  galvanometer  is  based  upon  the  principles  an- 
swered in  the  second  series  of  answers,  where  a  magnetic  needle  sus- 
pended (freely)  in  the  center  of  a  looped  or  coiled  conductor,  is  de- 
flected by  the  current  of  electricity  passing  around  the  coil  or  loop. 

In  ordinary  practice,  the  needle  is  suspended,  either  upon  a  point 
projecting  into  an  agate  cup  fixed  into  the  needle,  or  by  a  fiber  sus- 
pension. In  the  simpler  form  of  galvanometers,  the  magnetic  needle 
itself  swings  over  a  dial  graduated  in  degrees;  in  other  forms,  a  light 
index  needle  is  rigidly  attached  to  the  magnetic  needle  and  swings  over 
a  similar  dial. 

In  the  more  sensitive  galvanometers  a  small  reflecting  mirror  is 
attached  to  the  fiber  suspension  and  reflects  a  beam  of  light  upon  a 
horizontal  scale  situated  several  inches  from  the  galvanometer. 

In  any  of  these  galvanometers,  when  no  current  is  flowing  in  the 
coils,  the  needle  should  point  in  a  direction  parallel  to  the  length  of  the 
coil.  The  measuring  of  currents  by  most  galvanometers  depends  upon 
the  magnetic  needle  being  held  in  position  by  the  magnetic  attraction 
of  the  earth's  magnetism  or  the  attraction  of  some  adjacent  magnet. 

When  a  current  of  electricity  passes  around  the  coil,  the  tendency 
is  to  deflect  the  magnetic  needle  at  right  angles  to  its  original  posi- 
tion, while  the  tendency  of  the  earth's  magnetism  is  to  oppose  the 
movement.  The  couple  thereby  produced  will  cause  the  needle  to  be 
deflected  a  certain  number  of  degrees  from  its  original  position,  de- 
pending upon  the  relative  strength  of  the  two  magnetic  fields.  The 
stronger  the  current  in  the  coil  the  greater  the  deflection.  With  a 
galvanometer  of  standard  dimensions  and  a  magnetic  field  of  known 
strength,  such  as  the  earth's  magnetism  at  a  convenient  place  on  its 
surface,  a  strength  of  a  current  can  be  conveniently  adopted  as  a  unit 
which  will  produce  a  certain  deflection;  all  other  galvanometers  can 
be  calibrated  from  this  standard,  and  their  dials  graduated  to  read 
the  strength  of  currents  directly  in  the  conventional  unit  adopted. 

168 


Q.  45.   (1899-1900.)     What  is  the  ohm? 
How  can  electric  resistance  be  denned? 

What  is  one  of  the  most  important  quantities  in  electrical  measure- 
ments? 

What  is  meant  by  the  international  ohm? 

Ans.  45.    The  ohm  is  the  unit  of  resistance. 

It  has  been  previously  stated  that  the  resistance  varies  in  different 
substances;  that  is,  one  substance  offers  a  higher  resistance 
to  a  current  of  electricity  than  another.  Electrical  resistance  can, 
therefore,  be  denned  as  a  property  of  matter,  varying  with  different 
substances,  and  in  virtue  of  which  such  matter  opposes  or  resists  a  pas- 
sage of  electricity. 

The  resistance  which  all  substances  offer  to  the  passage  of  an 
electric  current  is  one  of  the  most  important  quantities  in  electrical 
measurement.  In  the  first  place,  it  is  that  which  determines  the 
strength  of  an  electric  current  in  any  circuit  in  which  a  difference  of 
potential  is  consequently  maintained,  as  shown  in  Ohm's  Law.  In  the 
second  place,  the  unit  of  resistance,  the  ohm,  is  the  only  unit  in  elec- 
trical measurements  for  which  a  material  standard  can  be  adopted, 
other  quantities  being  measured  by  the  effect  they  produce. 

The  basis  of  any  system  of  physical  measurements  is  generally  some 
material  standard,  conventionally  adopted  as  a  unit,  physical  meas- 
urements in  each  system  being  made  in  comparison  with  that  unit. 

The  unit  of  electrical  resistance  now  universally  adopted  is  called 
the  international  ohm. 

One  international  ohm  is  the  resistance  offered  by  a  column  of  pure 
mercury  106.3  centimeters  in  length,  1  square  millimeter  in  sectional 
area  at  32°  F.,  or  at  a  temperature  of  melting  ice. 

The  dimensions  of  the  column  expressed  in  inches  are  as  follows: 
Length,  41.85  inches;  sectional  area,  .00155  sq.  inch. 


Q.  46.  (1899-1900.)  What  is  said  of  the  resistance  of  conductors  at 
equal  temperatures,  irrespective  of  the  current  flowing  through  them, 
or  the  electro-motive  force  of  the  current? 

In  a  given  conductor,  which  offers  a  resistance  of  2  ohms  to  a  cur- 
rent of  1  ampere,  what  would  be  the  resistance  in  the  same  conductor 
if  12  amperes  were  flowing  through  it? 

Ans.  46.  The  resistance  of  a  given  conductor  at  equal  temperatures 
is  always  constant,  irrespective  of  the  strength  of  current  flowing 
through  it  or  the  electromotive  force  of  the  current. 

Hence,  if  a  given  conductor  offers  a  resistance  of  2  ohms  to  a  cur- 
rent of  1  ampere,  it  offers  the  same  amount  of  resistance,  no  more  nor 
less,  to  a  current  of  12  amperes. 


Q.  47.  (1899-1900.)  How  is  the  resistance  of  a  given  conductor  in- 
fluenced by  a  change  in  length? 

What  will  be  the  resistance  of  two  miles  of  copper  wire,  if  the  re- 
sistance of  ten  feet  of  the  same  wire  is  .013  ohms? 

Ans.  47.  If  the  length  of  a  conductor  be  doubled,  its  resistance  will 
be  doubled;  that  is,  the  resistance  of  a  given  conductor  increases  as 
the  length  of  the  conductor  increases,  the  resistance  being  directly  pro- 
portional to  the  length  of  the  conductor. 

When  it  is  required  to  find  the  resistance  of  a  conductor  of  which 
the  length  is  varied,  other  conditions  remain  unchanged,  the  follow- 
ing formula  may  be  used: 


169 


in  which 

rt  =  the  original  resistance; 
r2  =  the  required  resistance; 
h  =  the  original  length; 
12  =  the  changed  length. 

As  in  all  examples  of  proportion,  the  two  lengths  must  be  reduced 
to  the  same  unit. 

By  this  formula  we  see  that  the  resistance  of  a  conductor  after  its 
length  is  changed  is  equal  to  the  original  resistance  multiplied  by  the 
changed  length,  and  the  product  divided  by  the  original  length. 
In  two  miles  there  are  10,560  ft. 
Then  by  our  formula — 

.013  X  10560 

The  required  resistance  = =13.728  ohms. 

10 


Q.  48.  (1899-1900.)  How  is  the  resistance  of  a  given  conductor  in- 
fluenced by  a  change  of  the  area  of  the  conductor? 

If  the  resistance  of  a  conductor,  whose  sectional  area  is  .025  sq.  in. 
is  .32  ohms,  what  will  be  the  resistance  of  a  conductor  if  its  sectional 
area  is  increased  to  .25  sq.  in.?  How  does  the  resistance  of  a  round 
conductor  vary? 

The  resistance  of  a  round  copper  wire  .2  inches  diameter  is  45  ohms; 
what  will  be  the  resistance  of  the  same  kind  of  wire  .4  inches  in  diam- 
eter? What  is  said  of  the  resistance  of  two  or  more  conductors  in 
series? 

Ans.  48.  If  the  sectional  area  of  a  conductor  is  doubled,  the  re- 
sistance will  be  halved.  We  may,  then,  obtain"  the  value  of  the  re- 
sistance of  a  conductor  from  any  change  of  sectional  area  by  the  fol- 
lowing formula: 

rt  ai 

In  which— 

ri  =  the  original  resistance  of  the  conductor; 
r2  =  the  changed  resistance; 
a!  =  the  original  area; 
az  =  the  changed  area. 

From  the  relations  here  expressed  it  will  be  seen  that  the  resist- 
ance varies  inversely  as  the  sectional  area;  that  is,  the  resistance  of  a 
given  conductor  diminishes  as  its  sectional  area  increases. 

The  resistance  of  a  conductor  is  independent  of  the  shape  of  its 
cross-section. 

By  our  formula  the  conditions  as  stated — 
Sectional  area  =  .025  sq.  in.        I 

Resistance  =  .32   ohms.  changed  area  .25  sq.  m. 

.32  X  .025 

The  required  resistance  = =  .032  ohms. 

.25 

When  comparing  resistance  of  round  copper  wire  the  following  for- 
mula is  used: 

l^D1 

r2  = 

dz 
In  which— 

r!  =  the  original,  or  known,  resistance; 
r2  =  the  required  resistance; 
D  =  the  original  diameter; 
d  =  the  changed  diameter. 

This  formula  is  based  on  the  rule  that,  since  the  sectional  area  of  a 
round  conductor  is  proportional  to  the  square  of  its  diameter  (sectional 

170 


area  —  diameter  squared  multiplied  by  the  constant  .7854),  the  re- 
sistance of  a  round  conductor  is  inversely  proportional  to  the  square 
of  its  diameter. 

By  our  formula  conditions  as  stated  in  the  question— 

A  round  copper  wire  .2  inches  diameter; 

The  resistance  45  ohms; 

The  resistance  when  the  diameter  is  increased  to  .4  inches,  by  our 
formula  will  be  equal  to  the  original  resistance  multiplied  by  the  diam- 
eter squared,  divided  by  the  changed  diameter  squared. 

Thus— 

45  X  .22          45   X  .4 

Required  resistance  = = =  11.25  ohms. 

.4'  .16 

The  resistance  of  two  or  more  conductors  connected  in  series  is 
equal  to  the  sum  of  their  separate  resistances.  As  an  example,  if  four 
conductors  having  separate  resistances  7,  11,  13  and  19  ohms,  respec- 
tively, are  connected  in  series,  their  total,  or  joint,  resistance  will  be 
equal  to  the  sum  of  the  separate  resistances,  or  50  ohms. 


Q.  49.  (1899-1900.)  What  is  a  microhm?  For  what  was  it  de- 
vised? 

In  order  to  compare  resistances  of  different  substances  how  should 
the  dimensions  compare?  Why? 

What  metal  under  like  conditions  offers  the  least  resistance?  What 
other  metal  comes  next? 

Does  the  resistance  of  a  given  conductor  always  remain  constant? 
Why.'  Does  the  resistance  increase  or  decrease  as  the  temperature 
raises  or  lowers?  Kow  is  this,  in  comparison  with  liquids  and  carbons? 

Ans.  49.  The  microhm  is  a  unit  of  resistance  devised  to  facilitate 
calculations  and  measurements  of  exceedingly  small  resistances,  and  is 
equal  to  one-millionth  of  an  ohm  (1/1,000,000). 

In  order  to  compare  the  resistances  of  different  substances,  the  di- 
mensions of  the  pieces  to  be  measured  must  be  equal;  for,  by  changing 
its  dimensions,  a  good  conductor  may  be  made  to  offer  the  same  resist- 
ance as  an  inferior  one. 

Under  like  conditions,  annealed  silver  offers  the  least  resistance  of 
all  known  substances;  soft  annealed  copper  comes  next  on  the  list, 
and  then  follow  all  other  metals  and  conductors. 

The  resistance  of  a  given  conductor,  however,  is  not  always  con- 
stant, it  changes  with  the  temperature  of  the  conductor. 

In  all  metals  the  resistance  increases  as  the  temperature  rises. 

In  liquids  and  carbons  the  resistance  decreases  as  the  temperature 
rises. 

Q.  50.  (1899-1900.)  What  is  the  variation  of  the  resistance  of  a  con- 
ductor caused  by  a  change  of  temperature  of  1°  called? 

How  would  you  find  the  resistance  of  a  conductor,  after  its  tempera- 
ture had  risen,  knowing  its  original  resistance  and  the  number  of  de- 
grees rise  in  temperature,  other  conditions  remaining  the  same? 

The  resistance  of  a  piece  of  copper  wire  at  32°  is  40  ohms;  deter- 
mine its  resistance  when  its  temperature  is  74°  F.,  the  temperature  co- 
efficient for  copper  is  .002155. 

How  would  you  determine  the  resistance  if  the  temperature  is 
dropped,  other  conditions  remaining  the  same? 

If  the  original  resistance  of  a  German  silver  wire  is  16  ohms, 
determine  its  resistance  after  its  temperature  has  fallen  15°  F.  Tem- 
perature coefficient  of  German  silver  is  .000244. 

Ans.  50.  The  amount  of  variation  in  the  resistance*  caused  by  a 
change  of  temperature  for  one  degree  is  called  the  temperature  co- 
efficient. 

171 


These  co-efficients,  however,  hold  true  for  a  limited  change  of  tem- 
nprature  and  should  not  be  used  with  extreme  changes. 
P    ?o "find  the  resistance  of  a  conductor  after  its  temperature  has  risen, 
knowing  its  original  resistance  and  the  number  of  degrees  rise,  other 
condlSfns  remaining  unchanged,  multiply  the  original  resistance  by 
1  plus  the  product  of  the  number  of  degrees  rise  and  the  temperature 
co-efficient. 
Formula— 

r2  =  rx  (  1  +  t  k  ) 
In  which — 

ri  =  the  original  resistance; 

r2  =  the  resistance  after  the  temperature  change; 
t  =  rise  or  fall  of  temperature  in  degrees  F.; 
k  =  temperature  co-efficient. 
Resistance  of  copper  wire  at  32°  F.  is  40  ohms. 

At  a  temperature  of  74°  F.  (equals  42°  F.  rise)  the  temperature  co- 
efficient is  .002155. 

Then    according  to  our  rule,  or  formula — 

The  changed  resistance  =  40  (1  +  42  X  .002155)  =43.6204  ohms. 
To  determine  the  resistance  if  the  temperature  drops:    Divide  the 
original  resistance  by  1  plus  the  product  of  the  number  of  degrees  fall 
and  the  temp.  coef. 
Formula — 

r1 

1  +  tk 

The  letters  having  the  same  value  as  above. 

If  the  original  resistance  of  a  German  silver  wire  is  16  ohms;  the 
temp,  drops  15°  F.  the  temp.  coef.  of  German  silver  wire  is  .000244. 
By  our  rule  or  formula — 

16 

The  changed  resistance  = =  15.941  ohms. 

1  +  15  X  .000244 


Q.  51.  (1899-1900.)  What  is  understood  by  specific  resistance  of 
substances? 

What  instrument  is  usually  used  in  measuring  resistances?  Give  a 
brief  description  of  this  instrument. 

Ans.  51.  Specific  resistance  is  the  term  given  to  the  resistance  of 
substances  of  unit  length  and  unit  sectional  area  at  some  standard 
temperature. 

The  specific  resistance  of  a  substance  is  the  resistance  of  a  piece  of 
that  substance  one  inch  in  length  and  one  square  inch  in  sectional  area 
at  32°  F.;  that  is,  at  a  temp,  of  melting  ice;  this  may  also  oe  expressed 
as  the  resistance  of  a  cube  of  that  substance  taken  between  two  oppos- 
ing faces. 

A  list  of  the  common  metals  in  the  order  of  their  relative  resistances 
beginning  with  silver  as  offering  the  least  resistance: 

Silver,  annealed 1.000 

Silver,  hard  drawn  1.086 

Copper,  annealed   1.063 

Copper,  hard  drawn 1.086 

Etc.,  etc. 

An  instrument  known  as  "Wheatstone  Bridge"  is  usually  used  in 
laboratory  practice  in  determining  the  resistance  of  different  sub- 
stances, coils,  etc. 

These  are  different  forms  of  construction,  some  simple,  others  elab- 
orate, with  a. galvanometer  in  circuit. 

A  simple  form  in  laboratory  practice,  where  small  currents  are  used 
and  great  accuracy  is  required,  the  resistance  coils  are  enclosed  in  a 

172 


wooden  box  and  the  actual  resistance  of  each  coil  is  carefully  deter- 
mined, the  separate  coils  offering  resistances  from  one  ohm  upward  to 
5,000  ohms. 

The  operation  of  adjusting  the  resistance  is  by  the  means  of  removal 
plugs,  allowing  the  current  to  pass  through  all  or  part  of  the  coils,  as 
the  case  may  require. 


Q.  52.  (1899-1900.)  What  does  the  term  "volt"  denote?  What  three 
facts  are  to  be  carefully  noted  regarding  the  application  of  "Ohm's 
law"  to  closed  circuits? 

Ans.  52.    The  volt  is  the  practical  unit  of  electromotive  force. 

In  mechanics,  pressures  of  all  kinds  are  measured  by  the  effects 
they  produce;  similarly,  in  electrotechnics,  potential  is  measured  by 
the  effect  it  produces. 

By  definition  the  volt  is  that  electromotive  force  which  will  main- 
tain a  current  of  one  ampere  in  a  circuit  of  one  ohm's  resistance. 

With  a  known  resistance  in  ohms  and  a  known  strength  of  current 
in  amperes  the  electromotive  force  is  determined  by  "Ohm's"  law,  in 
which  the  current  in  amperes  multiplied  by  the  resistance  in  ohms 
equals  the  electromotive  force  in  volts. 

Or,  E  —  C  R. 

E  =  electromotive  force  in  volts. 

C  =  current  in  amperes. 

R  =  resistance  in  ohms. 

The  application  of  Ohm's  law  to  closed  circuits,  the  following  facts 
are  to  be  carefully  noted: 

The  current  (C)  is  the  same  in  all  parts  of  the  circuit,  except  in 
the  cases  of  derived  circuits,  where  the  sum  of  the  currents  in  the 
separate  branches  equal  to  the  current  in  the  main,  or  undivided, 
branches. 

The  resistance  (R)  is  the  resistance  of  the  internal  circuit  plus 
the  resistance  of  the  external  circuit. 


Q.  53.  (1899-1900.)  How  do  you  determine  the  strength  of  a  current 
in  amperes,  the  E.  M.  F.  and  R.  being  given? 

If  the  two  electrodes  of  a  simple  galvanic  cell  are  connected  by  a 
conductor  whose  resistance  is  1.5  ohms;  the  internal  resistance  of  the 
cell  is  6  ohms,  and  the  total  E.  M.  F.  developed  is  1.83  volts,  what  is 
the  strength  of  the  current  flowing  in  the  circuit? 

Ans.  53.  The  strength  of  the  current  (C)  in  amperes  divided  by 
the  resistance  (R)  in  ohms  equals  the  electromotive  force  (E)  in  volts. 

Internal   resistance    6.    ohms. 

External  resistance  1.5  ohms. 

Total  resistance  of 7.5  ohms. 

Then,  according  to  Ohm's  law — 
E       1.83 

C  =  —  = =  .244  amperes. 

R        7.5 
The  strength  of  the  current  flowing  in  the  circuit. 


Q.  54.  (1899-1900.)  How  would  you  find  the  total  resistance  in 
ohms  of  a  closed  circuit  when  the  E.  M.  F.  and  the  strength  of  the  cur- 
rent is  known? 

The  total  E.  M.  F.  developed  in  a  closed  circuit,  1.7  volts,  and  the 
current  is  .7  amperes;  find  the  resistance  in  ohms. 

Ans.  54.  The  total  resistance  in  ohms  of  a  closed  circuit  equals  the 
electromotive  force  (E)  in  volts  divided  by  the  strength  of  the  current 
(C)  in  amperes. 

173 


Electromotive  force  =  1.7  volts. 
Current  =  .7  amperes. 

1.7 
The  resistance  equals  =2.428  ohms. 

The  resistance  in  the  circuit. 


Q.  55.  (1899-1900.)  It  is  desired  to  transmit  a  current  of  15  am- 
peres to  a  receptive  device,  situated  1,000  feet  from  the  source;  the 
total  E  M  F.  generated  is  125  volts,  and  only  1-10  of  this  potential  is 
to  be  lost  in  the  conductors,  leading  to  and  from  the  device;  find  the 
resistance  of  the  two  conductors.  How  much  will  be  the  resistance 
per  foot  of  the  conductors? 

In  a  voltaic  cell,  what  is  to  be. understood  by  the  term,  available 
or  external  E.  M.  F.? 

The  internal  or  generated  E.  M.  F.? 

Ans.  55.  1/10  of  125  volts  =  12.5  volts,  which  represents  the  drop, 
or  loss  in  potential  on  the  two  conductors. 

Let— 

£'  =  12.5  volts. 

C  =  the  current  in  amps.  15. 

R1  =  total  resistance  of  the  two  conductors. 

Then — 

E1       12.5 

R1  =  —  = =  .8333  +  ohms. 

C          15 

The  resistance  per  foot  of  conductor  equals— 
.8333  + 

— r-  =  .0004166  +  ohms. 
2000 

The  difference  of  potential  between  the  two  electrodes  of  a  simple 
voltaic  cell  when  no  current  is  flowing;  that  is,  when  the  circuit  is 
open,  is  always  equal  to  the  total  E.  M.  F.  developed  in  the  cell,  and 
is  called  the  internal,  or  generated,  E.  M.  F. 

When  a  current  is  flowing;  that  is,  when  the  circuit  is  closed,  a 
certain  amount  of  the  potential  is  expended  in  forcing  the  current 
through  the  internal  resistance  of  the  cell  itself. 

Hence,  the  difference  of  potential  between  the  two  electrodes  when 
the  circuit  is  closed  is  always  smaller  than  when  the  circuit  is  open. 

This  difference  of  potential  between  the  two  electrodes  when  the 
circuit  is  closed,  is  called  the  available,  or  external,  E.  M.  F.,  to  dis- 
tinguish it  from  the  internal,  or  total,  generated  E.  M.  F. 


Q.  56.  (1899-1900.)  In  a  voltaic  cell,  the  total  generated  E.  M.  F. 
equals  2.5  volts,  and  the  internal  resistance  is  .7  ohms;  if  a  current  of 
1.5  amps,  flows  through  the  cell  when  the  circuit  is  closed,  what  is 
the  available  E.  M.  F.  developed  by  the  cell,  or  what  is  the  difference 
in  potential  between  the  two  electrodes? 

How  do  you  find  the  E.  M.  F.  (in  volts)  in  a  closed  circuit  when 
the  strength  of  the  current  and  the  total  resistance  is  known? 

Ans.  56.     To  find  the  available  E.  M.  F.  of  a  cell: 

Let— 

E  =  the  total  generated  E.  M.  F.; 

E^=the  available  E.  M.  F.  when  the  circuit  is  closed; 

—  the  current  in  amp.  flowing  when  the  circuit  is  closed; 

r,  =  the  internal  resistance  of  the  cell. 

Then  the  drop,  or  loss,  of  potential  in  the  cell  equals  C  r,. 

The  available  E.  M.  F.    E1  =  E  —  C  r^ 

174 


If  the  total  E.  M.  F.  equals  2.5  volts;  the  internal  resistance  equals 
7  ohms;  with  a  current  flowing  of  1.5  amps.,  the  available  E.  M.  F. 
will  equal— 

Ei  =  E  —  C  r,  =  2.5  — 1.5  X  .7  =  1.45  volts 

or  the  difference  in  potential  between  the  two  electrodes  when  the  cir- 
cuit is  closed. 

To  find  the  E.  M.  F.  (in  volts)  in  a  closed  circuit,  when  the  strength 
of  the  current  and  the  total  resistance  is  known: 

The  available  E.  M.  F.  of  a  cell  is  equal  to  the  difference  between 
the  total  generated  E.  M.  F.  and  the  potential  expended  in  forcing  the 
current  through  the  internal  resistance  of  the  cell  when  the  circuit  is 
closed. 

From  Ohm's  Law:  This  drop  of  potential  in  the  cell  itself  is  equal 
to  the  product  of  the  internal  resistance  in  ohms  and  the  strength  of 
the  current  in  amps,  flowing  through  the  circuit. 


Q.  57.  (1899-1900.)  The  internal  resistance  of  a  closed  circuit  is  3 
ohms,  and  the  external  resistance  is  4  ohms,  the  current  flowing  is  .5 
amperes;  what  is  the  E.  M.  F.  developed? 

What  is  understood  by  a  drop  of  potential?  How  is  this  drop 
usually  expressed? 

If  a  circuit  in  which  a  current  of  4  amperes  is  flowing  having  three 
separate  resistances  as  (in  diagram)  a  to  b,  b  to  c  ,and  c  to  d,  the 
resistance  from  a  to  b  1.6  ohms,  b  to  c  is  3.2  ohms  and  c  to  d  4.3  ohms, 
find  the  difference  in  potential  from  a  to  b,  b  to  c  and  c  to  d?  And  from 
a  to  d? 


Give  a  general  explanation  of  conductivity  in  regard  to  resistance? 

When  the  resistance  of  two  branches  are  unequal,  how  will  the  cur- 
rent divide  between  them  in  a  derived  circuit? 

Ans.  57— 

Let  R  ==  total  resistance  =  n  +  ^  =  .7  ohms. 

rl  =  internal  resistance;   3  ohms. 

r-2  =  external  resistance;   4  ohms. 

C  =  current  in  amps.  .5  amps. 

The  total  E.  M.  F.  =  C  R  =  .5  X  7  =  3.5  volts. 

The  drop  in  potential  =  Cr1  =  .5  X  3  =  1.5  volts. 

The  available  E.  M.  F.  =  3.5  —  1.5  =  2.0  volts. 

Drop  or  loss  of  Potential:  — 

By  referring  again  to  water  flowing  through  a  pipe;  though  the 
quantity  of  water  which  passes  is  the  same  at  any  cross  section  of  the 
pipe,  the  pressure  per  square  inch  is  not  the  same.  It  is  this  dif- 
ference of  pressure  that  causes  the  water  to  flow  between  two  points 
against  the  friction  of  the  pipes. 

This  is  precisely  similar  to  a  current  of  electricity  flowing  through  a 
conductor.  Though  the  quantity  of  electricity  that  flows  is  equal  at 
all  cross  sections,  the  E.  M.  F.  is  by  no  means  the  same  at  all  points 
along  the  conductor. 

It  suffers  a  loss  or  drop  of  electrical  potential  in  the  direction  in 
which  the  current  is  flowing,  and  it  is  this  difference  of  electrical  po- 
tential that  causes  the  electricity  to  flow  against  the  resistance  of  the 
conductor.  Ohm's  law  not  only  gives  the  strength  of  a  current  in  a 
closed  circuit,  but  also  the  difference  of  potential  in  volts  along  the 
circuit. 

The  difference  of  potential  (E1)  in  volts  between  any  two  points  is 
equal  to  the  product  of  the  strength  of  the  current  (C)  in  a.m.neres  and 

175 


the  resistance  (R1)  in  ohm's  of  that  part  of  the  circuit  between  those 
two  points,  or — 

B1  =  CR1 

E1  also  represents  the  loss  or  drop  of  potential  in  volts  between  the 
two  points. 

If  any  two  of  these  quantities  are  known  the  third  can  readily  be 
found,  for  by  transposing — 

E1  E1 

C  = and  R1  = . 

R1  C 

See  sketch—  R1  =  separate  resistances,  E1  =  the  drop  in  potential 
between  a  and  d,  according  to  Ohm's  law. 

E1  =  C  R1 

The  difference  of  potential  between — 
a  and  b  =  4  X  1.6  =   6.4  volts. 
D  and  c  =  4  X  3.2  =  12.8  volts, 
c  and  d  =  4  X  4.3  =  17.2  volts. 

aandd  =  -  36.4  volts. 

36.4  volts  represents  the  drop  of  potential  caused  by  a  current  of  4 
amperes  flowing  between  the  points  a  and  d. 

Conductivity   can   be   denned   as  the   facility   with   which   a  body 
transmits  electricity,  and  is  the  opposite  to  resistance. 
'     For  example,  copper  is  high  conductivity  with  low  resistance;  mer- 
cury is  high  resistance  and  low  conductivity. 

In  other  words,  conductivity  is  the  inverse  or  reciprocal  of  resist- 
ance. 

The  conductivity  of  any  conductor  is,  therefore,  unity  divided  by 
the  resistance  of  that  conductor;  and  conversely  the  resistance  of  any 
conductor  is  unity  divided  by  its  conductivity. 

When  the  current  flows  through  two  branches  of  a  derived  circuit 
and  the  resistances  are  equal  the  current  will  divide  equally  between 
the  two  branches. 

When  the  resistances  of  the  two  branches  are  unequal,  the  current 
will  divide  between  them  in  inverse  proportion  to  their  respective  re- 
sistances. 


i.  58.   (1899-1900.)     In  the  accompanying  diagram,  if  the  resistance 
2  ohms,  and  the  resistance  of  T2  =  4  ohms,  find  the  separate  cur- 


rent,  cf  and  "ca,  in  the  two  branches  respectively?  When  C=  (main 
current),  30  amperes,  in  the  undivided  mains,  a  and  b. 

If  the  separate  resistances  of  two  conductors  are  equal,  what  will 
be  their  joint  resistance  when  connected  in  parallel? 

Give  the  rule  when  the  separate  resistances  are  unequal? 

What  is  the  joint  resistance  when  three  or  more  conductors  are 
connected  in  parallel? 

fn  a  derived  circuit  of  any  number  of  branches,  what  would  be 
the  difference  of  potential  between  where  the  branches  divide  and 
where  they  unite? 

How -can  the  separate  resistances  of  the  branches  of  a  derived 
circuit  be  determined? 

176 


Ans.  58.     See  sketch: 
F!  =  2  ohms; 
r2  =  4   ohms; 
C  =  30  amperes; 

d  =  current  in  one  branch  of  the  derived  circuit; 
C2  =  current  in  the  other  branch  of  the  derived  circuit. 
The  resistances  in  the  two  branches  are  i\  and  r2,  therefore,  ct  :  ca 
: :  T2  :   r:. 

By  algebra  this  proportion  gives  the  two  following  formulas: 
Cr2          30  X  4       120 

For  the  first  branch  c,  = = = =  20    amps 

n  +  r2       2  +  4  6 

Cri         30  X  2         60 

For  the  second  branch  c2  = = =  — * —  =  10  amps 

rt  +  r2       2  +  4          6 

In  the  branch  ci  20  amperes  of  current  will  flow. 
In  the  branch  c2  10  amperes  of  current  will  flow. 
Reducing  our  formula  to  rule  we  have  for  the  first  branch. 
Of  two  branches  in  parallel,  dividing  from  a  main  circuit,  the  cur- 
rent in  the  first  branch  is  equal  to  the  current  in  the  main  multiplied 
by  the  resistance  of  the  second  branch,  the  product  divided  by  the 
sum  of  the  resistances  of  the  two  branches. 
For  the  second  branch, 

The  current  of  the  second  branch  is  equal  to  the  current  of  the 
main  circuit  multiplied  by  the  resistance  of  the  first  branch,  the 
product  divided  by  the  sum  of  the  resistances  of  the  two  branches. 

If  the  resistances  of  two  conductors  arc  equal,  their  joint  resistance 
when  connected  in  parallel  is  one-half  the  resistance  of  either  con- 
ductor. 

When  the  separate  resistances  of  two  conductors  in  parallel  are 
unequal,  the  determination  of  their  joint  resistance  when  connected 
in  parellel  involves  some  calculation. 

Referring  to  the  diagram,  the  conductivities  of  the  branches  are 
1  1 

—  and  — ,  respectively. 

F!  Ta 

Hence  their  joint  conductivity  when  connected  in  parallel  is 

1        1        r2  +  r, 


Now,  since  the  resistance  of  any  conductor  is  the  reciprocal  of  its 
conductivity,  then  the*  joint  resistance  of  the  two  branches  in  parallel 
is  the  reciprocal  of  their  joint  conductivity; 

r2  +  r1         FJ  r2 
or,  1  -r-  -  —  . 


Hence,  joint  resistance  R11  =  -  . 

Fj  +  F2 

That  is,  the  joint  resistance  of  two  conductors  connected  in  par- 
allel is  equal  to  the  product  of  their  separate  resistances  divided  by 
the  sum  of  their  separate  resistances. 

The  joint  resistance  of  three  or  more  conductors,  connected  in 
parallel,  is  equal  to  the  reciprocal  of  their  joint  conductivity. 

If  in  our  diagram  we  connect  in  another  conductor,  making  three 
instead  of  two  branches,  then  the  joint  resistance  of  the  three  branches 

F!  r2  Ta 

jn  parallel  R1U  =  --- 


r2  rs  +  ri  r2  +  FJ  r3 

177 


In  a  derived  circuit  of  any  number  of  branches,  the  difference  of 
potential  between  where  they  divide  and  where  they  unite  is  equal  to 
the  product  of  the  sum  of  the  currents  in  the  separate  branches  and 
their  joint  resistance  in  parallel. 

The  separate  currents  in  the  branches  of  a  derived  circuit  can  be 
determined  by  finding  the  difference  of  potential  between  where  the 
branches  divide  and  where  they  unite  again,  and  dividing  the  result 
by  the  separate  resistance  of  each  branch. 

The  separate  resistance  of  the  branches  of  a  derived  circuit  can  be 
determined  by  finding  the  difference  of  potential  between  where  the 
branches  divide  and  where  they  unite,  dividing  the  result  by  the  sep- 
arate currents  of  each  branch. 

Examples  of  the  three  conditions: 

First  condition — 

If  the  currents  in  the  three  branches  are  16,  8  and  4  amperes, 
respectively,  and  the  joint  resistance  is  2  6-7  ohms,  then  the  difference 
of  potential  between  a  and  b 

6  20 

=  (16  +  8  +  4)  2  —  =  28  X  —  =  80  volts 

7  7 
Second  condition — 

Assume  that  the  separate  resistances  of  the  three  branches  are,  re- 
spectively, 5,  10  and  20  ohms,  and  the  difference  of  potential  from  a 
to  b  is  80  volts.  Then  the  current  in  the  first  branch  is  equal  to  80 
^5  =  16  amps;  second  branch  to  80-^10  =  8  amps;  third  branch  to 
80  -=-20  =  4  amps. 

If  the  difference  of  potential  between  a  and  b  is  80  volts,  and  the 
currents  in  the  separate  branches  are  16,  8  and  4  amperes,  respectively, 
then  the  resistance  of  the  first  branch  is  80  -f-  16  =  5  ohms;  second 
branch  is  80 '  -=-  8  =  10  ohms;  third  branch  is  80  -f-  4  =  20  ohms. 


Q.  59.  (1899-1900.)  How  can  the  strength  of  an  electric  current  be 
defined? 

What  is  the  practical  unit  of  electrical  quantity,  and  what  does  it 
represent?  What  is  the  rule  for  calculating  the  quantity  of  electricity 
which  has  passed  in  a  circuit  in  a  given  time  when  the  strength  in 
amperes  is  known?  Explain,  briefly,  the  term  electrical  work? 

The  principle  of  the  conservation  of  energy  teaches  that  energy 
cannot  be  destroyed;  what  follows  in  regard  to  electrical  energy? 

How  would  you  find  the  amount  of  electrical  work  accomplished, 
in  joules,  during  a  given  time  in  any  circuit?  . '"- 

Ans.  59.  The  strength  of  an  electric  current  can  be  defined  as  a 
quantity  of  electricity  flowing  in  one  second. 

The  practical  unit  of  electricity  is  called  the  coulomb. 

The  coulomb  is  such  a  quantity  of  electricity  as  would  pass  in  one 
second  through  a  circuit  in  which  the  strength  of  the  current  is  one 
ampere. 

To  calculate  the  quantity  of  electricity  which  has  passed  in  a  cir- 
cuit in  a  given  time  when  the  strength  of  the  current  in  amperes 
is  known: 

Let  Q  =  the  quantity  in  coulombs; 

C  =  the  strength  of  current  in  amperes; 
t  =  the  time. 

Then  Q  =  Ct. 

If  any  two  of  these  quantities  are  known  the  third  can  be  readily 
found. 

Q  Q 

By  transposition,  C  =  — ,  or  t  =  — . 
t  C 

178 


Electrical  Work:  — 

When  an  electric  current  flows  from  a  higher  to  a  lower  potential, 
electrical  energy  is  expended,  and  work  is  done  by  the  current. 

The  principle  of  the  conservation  of  energy  teaches  that  energy 
can  never  be  destroyed;  it  follows,  therefore,  that  if  energy  has  to  be 
expended  in  forcing  a  quantity  of  electricity  against  a  certain  amount 
of  resistance,  the  equivalent  of  that  energy  must  be  transformed  into 
some  other  form.  This  other  form  is  usually  heat;  that  is,  when  a 
quantity  of  electricity  flows  against  the  resistance  of  a  conductor,  a 
certain  amount  of  electrical  energy  is  transformed  into  heat  energy. 
The  actual  amount  of  heat  developed  is  an  exact  equivalent  of  the 
work  done  in  overcoming  the  resistance  of  the  conductor,  and  varies 
directly  as  the  resistance. 

For  example,  take  two  wires,  the  resistance  of  one  being  twice  that 
of  the  other,  and  send  currents  of  equal  strength  through  each. 

The  amount  of  heat  developed  in  the  wire  of  higher  resistance  will 
be  twice  that  developed  in  the  wire  offering  the  lower  resistance. 

The  unit  used  to  express  the  amount  of  mechanical  work  done  is 
known  as  the  foot-pound. 

The  work  done,  in  raising  any  mass  through  any  height,  is  found 
by  multiplying  the  weight  of  the  body  lifted  by  the  vertical  height 
through  which  it  is  raised;  similarly,  the  practical  unit  of  electrical 
work  is  that  amount  accomplished  when  a  unit  quantity  of  electricity, 
one  coulomb,  flows  between  a  potential  of  one  volt. 

The  unit  of  electrical  work  is,  therefore,  the  volt-coulomb,  and  is 
called  the  joule. 

1  joule  — .7373  foot-pound. 

By  the  means  of  the  following  formulas,  we  may  find  directly  the 
amount  of  electrical  work  accomplished  in  joules  during  a  given  time 
in  any  circuit. 

Let— 

J  =  electrical  work  in  joules; 
C  =  current,  in  amps; 
t  =  time,  in  seconds; 
E  =  potential  or  E.  M.  F.,  of  circuit; 
R  =  resistance  of  circuit. 

When  the  E.  M.  F.  and  current  are  known, 
J  =  C  Et. 

When  the  current  and  resistance  are  known, 
J  =  C2  R  t. 

When  the  E.  M.  F.  and  resistance  are  known, 
E2t 

T 

R 

To  reduce  the  work  as  expressed  in  joules  to  foot-pounds — 
foot-pounds  =  Joules  X  .7373 


Q.  60.  (1899-1900.)  What  is  the  unit  of  power  or  rate  of  doing 
work  called?  Give  three  rules  for  computing  the  power  in  watts. 
Give  four  rules  for  determining  the  H.  P.  The  H.  P.  being  given,  how 
would  the  number  of  watts  be  determined?  Either  way,  how  would 
the  same  be  expressed  in  kilowatts? 

Ans.  60.  Power,  or  rate  of  doing  work,  is  found  by  dividing  the 
amount  of  work  done  by  the  time  required  to  do  it. 

In  mechanics,  the  unit  of  power  is  called  the  horse-power.     - 

In  electrotechnics,  the  unit  of  power  is  the  watt. 

It  is  found  by  dividing  the  amount  of  electrical  work  done  by  the 
time  required  to  do  it. 

179 


Let— 

E  =  E.  M.  F.  in  volts;      . 

Q  =  quantity  of  electricity  in  coulombs; 

C=:  current  in  amps; 


e,ectr,ca,  work,  J  =  C  E  t. 
CEt 


Rule-     The  power  in  watts  is  equal  to  the  strength  of  the  current 
in  amperes  multiplied  by  the  E.  M.  F.  in  volts. 

C2Rt 
W  =  -  =  C2R. 

Rule:     The  power  in  watts  is  equal  to  the  strength  of  the  current 
in  'amperes  squared  multiplied  by  the  resistance  in  ohms— 

E2t        E* 

W  =  -  =  —  . 

Rt         R 

The  power  in  watts  is  equal  to  the  quotient  arising  from  dividing 
the  E.  M.  F.  in  volts  squared  by  the  resistance  in  ohms. 

H  P  =  746  watts  or  1  watt  =  -  H  P. 

746 
W 
HP  =  -  . 

746 

To  express  the  rate  of  doing  work  (electrical)  in  horse-power  units, 
divide  the  number  of  watts  by  746. 

EC  C2  R  E2 

H  P  =  -  .     H  P  =  -  .     H  P  =  -  . 
746  746  746  R 

.     To  express  the  electrical  power  in  kilo-watts. 
1,000  watts  equal  1  kilo-watt. 

By  substituting  1,000  for  746  in  the  preceding  formulas  the  results 
obtained  will  be  in  kilo-watts. 


Q.  61.  (1900-01.)  When  is  an  electric  current  generated  in  a  con- 
ductor? 

Name  two  experiments  from  which  the  foregoing  principle  is  de- 
ducted. 

To  what  are  due  currents  generated  in  a  conductor  cutting  lines  of 
force  and  those  induced  in  a  coiled  conductor  by  a  change  in  the  num- 
ber of  lines  of  force? 

Give  brief  explanation. 

In  calculations  how  is  it  convenient  to  make  distinctions  between 
the  two  cases? 

In  these  explanations  what  caution  should  be  observed? 

Ans.  61.  An  electric  current  is  generated  in  a  conductor  when  that 
conductor  is  moved  across  a  magnetic  field,  so  as  to  cut  the  lines  of 
force  at  an  angle. 

If  a  coiled  conductor  be  straightened  out,  forming  one  long  conduc- 
tor, then  be  moved  across  the  magnetic  field  at  right  angles  to  the  lines 
of  force,  a  current  will  be  generated  in  the  circuit.  The  current,  how- 
ever, immediately  subsides  when  the  motion  ceases,  no  matter  whether 
the  conductor  is  in  the  magnetic  field  or  not 

Should  the  conductor  be  moved  in  the  magnetic  field  with  its  length 
parallel  to  the  lines  of  force,  no  current  will  be  generated  in  the  circuit. 

In  reality  currents  generated  in  a  conductor  cutting  lines  of  force 


and  those  induced  in  a  coiled  conductor  by  a  change  in  number  of  lines 
of  force  which  pass  through  the  coil,  are  due  to  the  same  movement. 

For  every  conductor  conveying  an  electric  current  forms  a  closed 
coil  and  every  line  of  force  is  a  complete  circuit  by  itself.  Consequently, 
where  any  part  of  a  closed  coil  is  cutting  lines  of  force  the  lines  of 
force  are  passing  through  the  coil  in  a  definite  direction  and  changing 
at  the  same  rate  as  the  cutting. 

For  example,  if  a  closed  coil  is  represented  by  a  heavy  loop  and  a 
light  loop  represents  four  lines  of  force.  When  the  two  closed  loops 
are  brought  together,  the  closed  coil  is  cut  at  one  place  by  four  lines 
of  force  and  at  the  same  time  the  number  of  lines  of  force  passing 
through  the  closed  coil  increases  from  nothing  to  four. 

In  calculations,  however,  it  is  convenient  to  make  distinctions  be- 
tween the  two  cases;  in  the  one  case,  to  consider  that  the  current  is 
generated  by  a  conductor  of  a  certain  length  cutting  lines  of  force  at 
right  angles;  and,  in  the  other  case,  to  consider  that  the  current  in  a 
closed  coil  is  induced  by  the  change  in  the  number  of  the  lines  of  force 
passing  through  the  coil. 

.  In  these  explanations  it  must  not  be  forgotten  that  an  electric  cur- 
rent is  the  result  of  a  difference  of  potential  or  electromotive  force. 
Consequently  it  is  not  actually  a  current  that  is  generated  in  the  mov- 
ing wire,  but  an  electromotive  force;  for,  in  all  of  the  previous  experi- 
ments in  which  currents  are  induced  or  generated  in  a  conductor  by  the 
lines  of  force,  if  the  circuit  is  opened  at  any  point,  no  current  will 
now,  but  the  electromotive  force  still  exists. 


Q.  62.  (1900-01.)  How  many  methods  are  known  of  producing  an 
electromotive  force  by -induction  in  a  coiled  conductor? 

Name  them. 

Explain  each  method. 

Ans.  62.  There  are  three  methods  of  producing  an  electromotive 
force  by  induction  in  a  coiled  conductor — i.  e., 

"a" — Electro-magnetic  induction. 

"b" — Self  induction. 

"c" — Mutual  induction. 

In  electro-magnetic  induction  the  change  in  the  number  of  lines  of 
force  which  pass  through  the  coil  is  due  to  some  relative  movement 
between  the  coil  and  a  magnetic  movement.  For  example,  by  thrusting 
a  magnet  into  the  coil  or  withdrawing  it;  or  again,  by  suddenly  thrust- 
ing the  coil  into  a  magnetic  field  with  its  plane  at  right  angles  to  the 
lines  of  force. 

In  self  induction  the  change  in  the  number  of  lines  of  force  is 
caused  by  sudden  changes  in  a  current  which  is  already  flowing  through 
the  coil  itself,  and  is  supplied  from  some  exterior  source. 

This  exterior  current  produces  a  magnetic  field  in  the  coil  itself, 
and  for  so  long  as  the  strength  of  the  current  remained  constant  there 
is  no  change  in  the  number  of  lines  of  force  which  pass  through  the 
coil. 

Should  the  strength  of  the  current  be  suddenly  increased,  a  change 
in  the  number  of  lines  of  force  occurs;  this  change  in  turn  induces 
an  electromotive  force  in  the  conductor,  which  opposes  the  original 
current  in  the  coil  and  tends  to  keep  the  current  from  rising. 

Its  action  is  similar  to  that  which  would  take  place  if  some  extra 
resistance  were  suddenly  inserted  into  the  circuit  at  the  instant  the 
strength  of  the  current  is  increased. 

The  original  current  eventually  reaches  its  maximum  strength  in 
the  coil  as  determined  by  Ohm's  law,  but  its  rise  is  not  instantaneous; 
it  is  greatly  retarded  by  this  induced  electromotive  force. 

If,  on  the  contrary,  the  strength  of  the  original  current  is  suddenly 
allowed  to  decrease,  another  change  is  produced  in  the  lines  of  force 
which  pass  through  the  conductor  or  coil;  this  new  change  induces  an 
electromotive  force  which  acts  in  the  same  direction  as  that  of  the 
original  current  and  tends  to  keep  it  from  falling. 

181 


As  in  the  previous  case,  however,  the  original  current  will  eventual- 
ly drop  to  its  maximum  strength,  as  determined  by  Ohm's  law,  but  will 
:all  gradually  and  a  fraction  of  a  second  will  elapse  before  it  becomes 
constant. 

In  short,  the  current  flowing  through  a  coiled  conductor  acts  as 
possessing  inertia;  any  sudden  change  in  the  strength  of  the  current 
produces  a  corresponding  electromotive  force  which  tends  to  oppose 
that  change  and  keep  the  current  at  a  constant  strength. 

Mutual  Induction. 


In  mutual  induction  two  separate  coiled  conductors,  one  conveying 
a  current  of  electricity,  are  placed  near  each  other,  so  that  the  mag- 
netic circuit  produced  by  the  one  in  which  the  current  is  flowing  is 
enclosed  by  the  other,  as  shown  by  the  sketch,  where  the  current  cir- 
culates around  the  coil  P  when  the  circuit  is  closed  at  b.  The  coil  P 
is  called  the  primary  or  exciting  coil,  and  the'  coil  S  is  called  the  sec- 
ondary coil. 

Any  sudden  change  in  the  strength  of  the  current  in  the  primary 
coil,  as,  for  instance,  breaking  the  circuit  at  b,  a  corresponding  change 
in  the  number  of  lines  of  force  in  the  magnetic  circuit  which  passes 
through  both  coils;  and  hence,  an  electromotive  force  is  induced  in  the 
secondary  coil. 

If  the  primary  circuit  is  completed  at  b,  and  the  current  tends  to 
rise  in  the  coil,  the  electromotive  force  induced  in  the  secondary  coil, 
causes  a  current  to  circulate  around  it  in  the  opposite  direction  to  the 
current  in  the  primary  coil. 

If,  on  the  contrary,  the  circuit  at  b  is  suddenly  broken  and  the 
current  in  the  primary  decreases,  the  induced  electromotive  force  in 
the  secondary  coil  causes  a  current  to  circulate  around  it  in  the  same 
direction  as  the  current  in  the  primary  coil. 

The  directions  of  an  induced  current  in  a  coil  depends  upon  the 
direction  of  the  lines  of  force  in  the  coil  and  whether  their  number  is 
increasing  or  diminishing.  If,  then,  two  facts  are  known,  the  direction 
in  which  the  current  circulates  around  the  coil  is  determined  by  the 
following  rule: 

Rule.— If  the  effect  of  the  action  is  to  diminish  the  number  of  lines 
of  force  that  pass  through  the  coil,  the  current  will  circulate  around 
the  coil  from  left  to  right,  or  clockwise,  as  viewed  by  a  person  looking 
along  the  magnetic  field  in  the  direction  of  the  lines  of  force;  but 
t  the  effect  is  to  increase  the  number  of  lines  of  force  that  pass  through 
the  coil,  the  current  will  circulate  around  in  the  opposite  direction 


Q.  63.  (1900-01.)  Give  a  convenient  method  for  remembering  the 
direction  of  a  current  generated  in  a  straight  conductor  when  the  con- 
ductor is  moved  in  a  magnetic  field  at  right  angles  to  the  lines  of  force. 
•  *  ti^Uvmm,ary  of  electr°-magnetic  induction  experiments,  what  law 

IS    6St3.DllSn.6Q? 

th/H^t6!'    1*ule--place  the  thumb,  forefinger  and  middle  finger  of 
the  right  hand  so  that  each  will  be  perpendicular  to  the  other  two;  if 


182 


the  forefinger  points  in  the  direction  of  the  lines  of  force,  and  the 
thumb  points  in  the  direction  in  which  the  conductor  is  moving,  then 
the  middle  finger  will  point  in  the  direction  toward  which  the  current 
generated  in  the  conductor  tends  to  flow. 

The  summary  of  these  electro-magnetic  induction  experiments 
can  be  stated  as  follows:  Electromotive  forces  are  generated  in  a 
conductor  moving  in  a  magnetic  field  at  right  angles  to  the  direction 
of  the  lines  of  force,  or  are  induced  in  a  coiled  conductor  when  a 
change  occurs  in  the  number  of  lines  of  force  which  pass  through  the 
coil.  

Q.  64.  (1900-01.)  To  what  is  E.  M.  F.  generated  in  a  moving  con- 
ductor cutting  lines  of  force  at  right  angles,  directly  proportioned? 

Can  you  explain  this  principle  by  the  use  of  cross-section  paper? 

Ans.  64.  The  E.  M.  F.  generated  in  a  moving  conductor  cutting 
lines  of  force  at  right  angles  is  directly  proportioned  to  the  rate  of 
cutting. 

If,  for  an  example,  that  a  magnetic  field  contains  100,000  lines  of 
force  and  a  conductor  is  moved  across  the  field  at  right  angles  in  such 
a  manner  as  to  cut  every  line  of  force  in  one  second,  then  the  rate  of 
cutting  is  100,000  lines  per  second;  if  it  occupied  two  seconds  of  time, 
then  the  rate  of  cutting  would  be  50,000  per  second,  and  the  electro- 
motive force  in  this  case  would  be  one-half  that  of  the  former  case. 


Q.  65.  (1900-01.)  Explain  the  change  in  direction  of  the  current  in 
a  coiled  conductor  moving  in  a  magnetic  field,  and  give  value  of  this 
current  at  different  points  in  one  revolution. 

Ans.  65.  When  the  coil  begins  to  revolve  in  the  magnetic  field,  a 
feeble  E.  M.  j?.  is  generated;  this  E.  M.  F.  causes  a  corresponding  cur- 
rent to  flow  through  the  circuit  in  a  positive  direction;  as  the  E.  M.  F. 
becomes  larger,  the  strength  of  the  current  in  the  circuit  becomes 
greater,  and  vice  versa. 

After  the  coil  is  rotated  one-half  of  a  revolution  and  the  direction 
in  which  the  E.  M.  F.  tends  to  act  becomes  negative,  the  direction  of 
tne  current  in  the  circuit  is  also  reversed. 

If  there  is  no  self  induction  to  retard  the  rise  and  fall  of  the  cur- 
rent in  the  circuit  the  strength  of  the  current  in  the  circuit  at  any 
instant  is  exactly  proportional  to  the  E.  M.  F.  that  is  being  generated 
in  the  coil  at  that  moment;  for,  according  to  "Ohm's"  law,  the  strength 
of  a  current  in  any  circuit  is  equal  to  the  E.  M.  F.  generated  in  that 
circuit,  divided  by  its  resistance. 

The  rising  and  falling  and  also  the  reversing  of  the  current  in  all 
parts  of  the  circuit  can  be  graphically  represented  on  cross-section 
paper. 

In  which  the  current  at  the  starting  point  in  each  revolution  is 
zero,  rising  to  the  maximum  strength  at  one-quarter  of  the  revolution, 
following  again  to  zero  at  the  one-half  point  of  the  revolution,  at  which 
point  the  current  is  reversed  and  again  rising  to  its  maximum  strength 
at  the  three-quarter  revolution,  falling  again  to  zero  strength  at  the 
completion  of  the  revolution,  when,  as  at  the  one-half  revolution  or 
zero  point,  the  current  is  reversed,  showing  that  the  strength  of  the 
current  is  at  its  maximum  strength  and  at  zero  strength  twice  in  each 
revolution.  At  intermediate  points  the  strength  of  the  current  is  pro- 
portional to  the  time  required  for  making  the  full  revolution. 


Q.  66.   (1900-01.)     Describe  an  alternating  current. 
Show  by  the  use  of  cross-section  paper  the  value  of  the  E.  M.  F.  in 
one  complete  revolution. 

What  is  understood  by  the  term  "alternation"? 
What  is  understood  by  the  term  "frequency"? 

183 


What  is  understood  by  the  term  "cycle"? 
A  "cycle"  represents  how  many  degrees? 
How  in  regard  to  time? 


x 

' 

^~^ 

\ 

/ 

\ 

A 

-B 

C 

\ 

D 

^ 

^ 

..  

„ 

Lx 

A  Time 

DESCRIPTION    OF    SKETCH. 

Ans.  66.  The  E.  M.  F.  that  is  generated  in  a  coil  at  every  instant 
during  one  complete  revolution  is  graphically  shown  in  the  sketch  by 
the  use  of  cross-section  paper.'  (See  sketch.) 

The  sum  of  the  divisions  between  A  and  E  represents  the  time  occu- 
pied by  the  coil  in  making  one  complete  revolution;  the  divisions  be- 
tween A  and  Y  represent  the  E.  M.  F.  which  tends  to  send  a  current 
through  the  coil  in  one  direction  during  the  first  half  of  the  revolution, 
and  the  divisions  between  A  and  ^  represent  the  E.  M.  F.  which  tends 
to  send  a  current  through  the  coil  in  an  opposite  direction  as  in  the 
last  half  of  the  revolution.  The  divisions  between  the  curved  line  and 
the  base  line  AE  give  the  E.  M.  F.  that  is  being  generated  in  the  coil 
at  any  instant  during  the  revolution,  and  the  direction  in  which  the 
E.  M.  F.  tends  to  act  depends  upon  whether  this  E.  M.  F.  is  above  or 
below  the  base  line  AE. 

For  convenience,  let  the  direction  of  the  E.  M.  F.  in  the  first  part 
of  the  revolution  be  called  positive  and  in  the  last  half  negative. 

For  example,  the  E.  M.  F.  that  is  generated  in  the  coil  when  it  has 
revolved  three-quarters  of  a  revolution  is  represented  by  the  distance 
between  D  and  the  curved  line,  which  in  this  case  is  two  divisions; 
and,  since  these  divisions  are  below  the  base  line,  the  direction  in 
which  this  E.  M.  F.  tends  to  act  is  negative. 

The  time  of  one  complete  revolution  is  represented  with  sub-divi- 
sions A,  B,  C,  D,  E. 

A  B  representing  one-quarter  revolution. 

A.  C.  representing  one-half  revolution. 

A  D  representing  three-quarter  revolution. 

A  E  representing  complete  revolution. 

The  electric  current  as  described  in  answer  65  is  an  alternating  cur- 
rent. 

The  term  Alternation.  In  each  reversal  of  current — that  is,  each  in- 
crease of  current  from  zero  to  maximum  and  the  decrease  to  zero 
again  is  called  an  alternation,  thus  two  alternations  occurring  in  one 
revolution. 

This  pair  of  alternations  is  called  a  cycle,  and  the  number  of  cycles 
that  occur  in  one  second  is  termed  the  frequency. 

A  cycle  represents  360°,  regardless  of  time. 


Q.  67.  (1900-01.)     What  is  meant  by  the  term  "commuting"? 
For  what  purpose  is  a  commutator  used? 
What  is  a  "pulsating"  current? 

How  can  the  strength  of  such  current  be  made  more  uniform  and 
pulsations  less  noticeable? 

magnet?  ^  ^  *&™*'  °f  inserting  an  iron  core  between  the  poles  of  a 
184 


Ans.  67.  The  term  commuting  means  to  change,  and  for  the  subject 
which  we  are  treating  would  mean  to  change  an  alternating  current 
into  a  continuous  current,  for  which  purpose  a  commutator  is  used.  A 
combination  of  two  segments  of  copper  insulated  apart  from  each  other, 
or,  in  fact,  any  number  of  segments  constitute  what  is  called  a  com- 
mutator. 

Two  or  more  copper  or  carbon  brushes  press  against  the  segments 
and  are  held  in  the  proper  position  while  the  coil  is  rotated. 

The  brushes  rub  or  brush  against  the  segments  and  make  electrical 
contact  only.  By  this  arrangement  the  current  in  the  external  circuit 
flows  continually  in  the  same  direction,  while  the  current  in  the  coil 
flows  in  two  directions  during  every  revolution.  But  the  strength  of 
the  current  in  the  external  circuit  is  by  no  means  constant.  It  rises 
from  zero  to  maximum  and  falls  again  to  zero  twice  in  every  revolu- 
tion, but  always  in  the  same  direction.  This  effect  is  produced  con- 
tinually in  the  external  circuit  if  the  coil  is  rotated  at  a  constant  speed. 
Tnese  impulses  in  the  strength  of  the  current  give  it  the  name  of 
"pulsating  current." 

The  strength  of  such  currents  can  be  made  more  uniform  and  the 
pulsations  less  noticeable  by  using  more  coils  connected  to  segments 
in  the  commutator,  the  planes  of  the  coils  being  placed  at  equal  angles 
from  each  other. 

In  last  year's  treatment  of  this  subject  the  term  permeability  was 
denned  as  the  facility  afforded  by  any  substance  to  the  passage  through 
it  of  lines 'of  force,  and  that  permeability  of  soft  iron  may  be  in  the 
ratio  of  2,000  to  1  as  compared  to  air;  and  that  when  a  magnetic  sub- 
stance is  brought  into  a  magnetic  field  the  lines  of  force  in  the  field 
crowd  together  and  all  try  to  pass  through  that  substance,  etc. 

Hence,  if  the  coils  are  wound  around  a  cylindrical  drum  of  iron  the 
number  of  lines  of  force  passing  through  the  coils  is  increased. 

These  coils  are  entirely  insulated  from  the  iron  core  by  some  non- 
conducting material,  otherwise  they  would  be  short  circuited  on  the 
core — that  is,  the  current  would  pass  through  the  iron  instead  of  pass- 
ing into  external  circuit. 

An  iron  core  inserted  between  the  poles  of  a  magnet  not  only  in- 
creases the  lines  of  force  from  the  magnet,  but  attracts  nearly  all  the 
stray  lines  of  force  from  the  surrounding  air. 


Q.  68.  (1900-01.)  In  a  rotated  coil,  where  will  the  greatest  differ- 
ence in  potential  be  found? 

What  means  are  used  to  utilize  this  difference  of  potential  between 
each  pair  of  wires? 

it  a  comparatively  large  number  of  turns  and  segments  are  used, 
what  will  be  the  effect  of  the  current?  Why? 

Ans.  68.  In  a  rotated  coil  the  greatest  difference  in  potential  will 
be  found  between  any  two  turns  diametrically  opposite  one  another 
when  they  pass  through  the  vertical  diameter.  To  utilize  this  difference 
ol  potential  between  each  pair  of  turns  as  they  arrive  in  a  .vertical 
position. 

This  is  accomplished  by  connecting  each  turn  to  a  separate  seg- 
ment of  a  commutator  by  a  small  conductor  and  allowing  two  brushes 
to  rub  against  the  commutator  at  two  points  diametrically  opposite  each 
other  on  the  vertical  diameter. 

If  a  comparatively  large  number  of  turns  and  segments  are  used 
the  current  flowing  through  the  external  circuit  will  practically  be  con- 
tinuous— that  is,  non-pulsating;  the  fluctuations  caused  by  the  brushes 
passing  from  one  segment  to  another  are  extremely  minute  and  pro- 
duce no  appreciable  change  in  the  strength  of  the  current  in  the  ex- 
ternal circuit. 

185 


Q.  69.  (1900-01.)     Describe  a  closed  coil  winding. 

Describe  an  open  coil  winding. 

What  is  meant  by  neutral  spaces? 

What  is  meant  by  neutral  points? 

Explain  shifting  of  brushes,  etc. 

Ans.  69.  Closed  Coil  Winding.  A  conductor  wound  upon  a  core  in 
this  manner  is  termed  a  "closed"  coil  winding,  since  all  the  turns  are 
connected  together  in  one  continuous  or  closed  coil  and  the  current  is 
obtained  from  it  by  lapping  into  each  turn  or  set  of  turns. 

In  the  case  where  the  turns  or  set  of  turns  are  separate  and  distinct 
from  each  other  and  their  ends  are  connected  to  opposite  segments  of 
a  commutator  the  winding  is  termed  an  "open"  coil  winding. 

The  parts  of  the  core  where  the  conductors  are  not  cutting  lines  of 
force  as  the  core  is  rotated  are  called  "neutral  spaces." 

The  two  opposite  parts  of  the  commutator  to  which  the  coils  are 
connected  are  called  the  neutral  point  of  the  commutator.  ...  - 

Each  individual  conductor  in  a  bipolar  machine  becomes  inactive 
twice  in  each  revolution  and  passes  through  two  neutral  spaces;  but 
this  fact  does  not  change  the  position  of  the  neutral  spaces — they  lie 
on  an  imaginary  diameter  approximately  perpendicular  to  the  lines  of 
force. 

This  same  effect  takes  place  in  the  commutator — i.  e.,  each  segment 
passes  through  two  neutral  spaces  during  one  revolution,  but  the  neu- 
tral points  remain  in  a  fixed  position  relative  to  the  neutral  spaces  of 
the  core. 

The  neutral  segments  of  a  commutator,  at  any  instant  during  a 
revolution,  are  those  segments  connected  to  the  conductors  passing 
through  two  neutral  spaces  at  that  instant. 

The  neutral  points  can  be  shifted  to  different  points  around  the 
commutator  by  changing  the  leads  from  the  coil  to  the  segment. 

In  order  to  collect  any  current  from  the  commutator  the  brushes 
must  be  at  the  neutral  points. 

The  current  flowing  through  the  winding  divides  at  the  neutral 
space  and  flows  through  the  coil  in  opposite  directions,  uniting  again 
at  the  other  neutral  space. 


Q.  70.  (1900-01.)  Suppose  a  magnetic  field  contains  25,000,000  lines 
of  force  and  a  conductor  cuts  the  total  number  in  the  same  direction  in 
one  second.  What  will  be  the  generated  E.  M.  F.? 

Suppose  there  were  25  conductors  connected  in  series,  cutting  the 
lines  of  force  at  equal  rates  (as  above),  what  will  be  the  generated 
E.  M.  F.? 

Suppose  in  the  latter  case  that  the  25  conductors  were  moved  across 
this  same  magnetic  field  at  the  rate  of  40  times  per  second,  wnat  would 
be  the  generated  E.  M.  F.? 

What  application  of  formula  can  be  made  in  relation  to  a  closed 
coil  conductor,  wound  upon  a  ring  or  drum  core? 

Ans.  70.  In  the  case  of  a  single  conductor  moving  across  a  mag- 
netic field  in  which  the  total  lines  of  force  is  known,  the  rate  of  cut- 
ting is  equal  to  the  total  number  of  lines  of  force  cut  by  the  conductor 
divided  by  the  time  required  to  cut  them. 

This  may  be  expressed  in  the  form  of  an  equation,  thus  the  rate 

of  cutting  =  — ,  where  N  =  the  total  lines  of  force  cut  and   t  =  the 

time  required  to  cut  them. 

By  definition:  One  volt  is  that  E.  M.  F.  generated  in  a  conductor 
when  it  is  cutting  lines  of  force  at  the  rate  of  one  hundred  million 
(100,000,000)  per  second. 

186 


Hence  E  =  -  when  E  is  the  E.  M.  F.  in  volts;   t  =  the  time  in 

10<t 
seconds  ;    100,000,000  =  108. 

For  example,  a  magnetic  field  contains  25,000,000  lines  of  force,  and 
the  conductor  cuts  the  total  number  in  one  direction  in  one  second, 
according  to  our  formula. 

25,000,000 
The  generated  E.  M.  F.  =  -  =.25  volt. 

100,000,000  X  1 

If  there  were  25  conductors  moving  across  the  magnetic  field  under 
the  same  conditions  as  in  previous  question,  then  will  the  generated 

25,000,000  X  25 
E.  M.  F.  =  ---  =6.25  volts. 

100,000,000  X  1 

If  the  25  conductors  were  moved  across  the  same  magnetic  field  40 
times  per  second, 
the  generated  E.  M.  F. 

An  equation  representing  this  operation  is 

(Nnt)  S  NSn 

E  =  -  =  eliminating  "t,"  E  =  --  ; 

10>t  108 

In  which— 

N  =  number  of  lines  of  force. 

n  =  number  of  times  per  sec.  one  conductor  cuts  the  lines  of  force. 
t  =  time  in  seconds. 
S  =  number  of  conductors. 

This  equation  can  be  applied  with  some  modifications  to  the  closed 
coil  conductor  wound  upon  either  the  ring  or  drum  core. 

In  case  of  the  ring  core,  E  in  the  equation  represents  the  maximum 
E.  M.  F.  in  volts  that  is  obtained  from  the  positive  and  negative  brushes 
when  the  core  is  revolved. 

N  is  the  total  number  of  lines  of  force  passing  from  the  north  pole 
through  the  core  to  the  south  pole  (bi-polar  dynamos). 

Each  wire,  therefore,  on  the  periphery  of  the  core,  cuts  the  total 
number  of  lines  of  force  twice  in  each  revolution,  or,  in  other  words, 
outside  wire  cuts  2N  lines  of  force  per  revolution. 

S  represents  the  number  of  outside  wires  on  the  periphery  through 
which  the  current  flows  in  series,  and  n  is  the  number  of  complete 
revolutions  per  second  of  the  core. 

Therefore,  the  maximum  E.  M.  F.  in  volts  that  is  obtained  from  the 
brushes  is  found  by  the  formula  — 
2NSn 


10s 

This  formula  holds  equally  true  for  the  drum  core. 

In  both  cases  the  number  of  outside  wires  through  which  the  current 
flows  in  series  is  equal  to  one-half  of  the  total  number  of  outside  wires. 
Hence,  by  using  the  same  magnetic  field  and  rotating  the  cores  at  equal 
speed,  the  E.  M.  F.  generated  in  both  cases  will  be  equal. 

The  difference  of  potential  between  the  brushes  when  the  external 
circuit  is  closed  is  somewhat  smaller  than  where  no  current  is  flowing; 
the  same  as  in  the  case  of  the  voltaic  cell,  a  part  of  the  total  E.  M. 
F.  developed  is  required  to  overcome  the  internal  resistance. 


Q.  71.   (1900-01.)     The  foregoing  questions  treat  upon  the  elemen- 
tary principle  and  physical  theory  of  a  dynamo. 
What  is  a  dynamo? 

What  are  its  three  most  essential  features? 
In  all  dynamos,  how  is  the  magnetic  field  produced? 

187 


Ans.  71.  A  dynamo  is  a  machine  for  converting  mechanical  energy 
into  electrical  energy  by  electro-magnetic  induction. 

The  three  essential  features  are: 

First,  a  magnetic  field. 

Second,  a  conductor  or  several  conductors,  called  an  armature,  in 
which  the  electromotive  force  is  generated  by  some  movement  relative 
to  the  lines  of  force  in  the  magnetic  field. 

Third,  a  commutator  or  collector  from  which  the  current  is  col- 
lected by  two  or  more  conducting  brushes. 

In  all  dynamos  the  magnetic  field  is  produced  either  by  a  perma- 
nent magnet  or  by  an  electro  magnet,  and  they  are  classified  accord- 
ingly. 

For  our  purpose,  however,  it  is  sufficient  to  consider  only  the  uni- 
form magnetic  field  lying  between  the  poles  of  some  large  magnet. 


Q.  72.  (1900-01.)  Give  general  rule  for  a  conductor  conveying  an 
electric  current  when  placed  in  a  magnetic  field. 

Give  a  convenient  rule  for  remembering  the  direction  of  motion 
imparted  to  a  conductor  conveying  an  electric  current  when  placed  in 
a  magnetic  field. 

If  a  vertical  conductor  in  which  a  current  is  flowing  downward, 
is  placed  in  front  of  the  north  pole  of  a  magnet,  in  which  direction 
will  the  conductor  tend  to  move? 

Ans.  72.  When  a  conductor  conveying  an  electric  current  is  placed 
in  a  magnetic  field,  the  conductor  will  tend  to  move  in  a  definite  direc- 
tion and  with  a  certain  force,  depending  upon  the  strength  and  direc- 
tion of  the  current  and  upon  the  direction  and  density  of  the  lines  of 
force  in  that  field. 

Rule. — Place  thumb,  forefinger  and  middle  finger  of  the  left  hand 
each  at  right  angles  to  the  other  two;  if  the  forefinger  points  in  the 
direction  of  the  lines  of  force  and  the  middle  finger  points  in  the  direc- 
tion toward  which  the  current  flows,  then  the  thumb  will  point  in  the- 
direction  of  movement  imparted  to  the  conductor. 

In  the  case  of  a  vertical  conductor  in  which  the  current  is  flowing 
downward  and  is  placed  in  front  of  the  north  pole  of  a  magnet  the 
conductor  will  tend  to  move  from  left  to  right. 


Q.  .73.  (1900-01.)  Compare  rule  given  in  question  63  and  the  rule 
given  in  question  72  and  explain  the  opposition  seemingly  of  one  to 
the  other  in  the  direction  of  current. 

Explain  the  theory  of  counter  torque  of  a  dynamo. 

Ans.  73.  Comparing  the  previous  rules  with  the  one  just  given,  it 
will  be  seen  that  the  two  appear  to  oppose  each  other;  or,  in  other 
words,  the  current  which  flows  in  the  former  case,  according  to  the 
latter  rule,  tends  to  oppose  the  motion  of  the  conductor  and  move  it  in 
the  opposite  direction. 

This  is  exactly  what  takes  place. 

When  a  conductor  is  moved  across  the  lines  of  force,  an  electric 
current  is  generated,  which  tends  to  send  a  current  in  a  definite  direc- 
tion. 

If  the  circuit  is  open  and  no  current  flows,  it  requires  no  force  to 
move  the  conductor  across  the  field;  but  if  the  circuit  is  closed  and  a 
current  flows  through  the  conductor,  then  the  action  of  the  lines  of 
force  on  the  current  opposes  the  original  motion  and  tends  to  stop  or 
retard  the  conductor. 

The  opposing  force  is  proportional  to  the  strength  of  the  current 
flowing  in  the  conductor.  But,  so  long  as  the  conductor  is  moved,  the 
applied  force  is  always  larger  than  the  counter  force. 

Hence  the  stronger  the  current  in  the  conductor  the  greater  will  be 
the  force  necessary  to  keep  the  conductor  moving  in  the  original  direc- 
tion. The  counter  force  would  never  actually  move  the  conductor  in 

1 88 


its  direction,  but  it  exerts  a  dragging  effect  upon  the  conductor  which 
would  reduce  its  speed  and  almost  stop  its  motion,  if  the  exterior  mo- 
tive force  is  not  increased. 

The  above  principles  explains  the  action  of  converting  the  mechan- 
ical energy  into  electrical  energy  in  a  dynamo. 

If  an  armature  is  properly  wound  and  connected  to  a  commutator, 
an  electromotive  force  is  generated  in  the  outside  conductors  on  the 
core,  causing  a  difference  of  potential  between  the  brushes.  If  the 
brushes  are  not  connected  to  an  external  circuit  and  no  current  is  flow- 
ing through  the  armature,  it  requires  no  energy  to  rotate  the  armature, 
excepting  a  small  amount  to  overcome  the  mechanical  friction  and  the 
loss  in  the  armature  iron  by  eddy  currents. 

If,  however,  connected  to  an  external  circuit  and  a  current  flowing 
through  the  armature  the  conditions  are  changed. 

The  lines  of  force  react  upon  the  current  in  the  conductors,  tending 
to  rotate  the  core  in  an  opposite  direction  and  to  retard  its  motion; 
the  stronger  the  current  the  greater  the  retarding  effect. 

Hence,  to  keep  the  speed  constant  and  to  generate  a  constant  B.  M. 
F.,  more  energy  must  be  supplied. 

This  retarding  effect  of  the  current  is  known  as  the  "Counter 
Torque"  of  a  dynamo. 


Q.  74.  (1900-01.)  Can  it  be  mathematically  proven  that  the  me- 
chanical energy  delivered  to  the  armature  from  any  exterior  source  is 
exactly  equal  to  the  electrical  energy  obtained  from  the  armature  plus 
tne  energy  lost  in  mechanical  friction,  eddy  currents  in  the  iron  and 
other  small  losses? 

What  are  the  losses  that  occur  in  an  armature? 

How  can  the  effect  of  armature  reaction  be  almost  entirely  elimi- 
nated? 

Ans.  74.  It  can  be  mathematically  proven  that  the  mechanical  en- 
ergy delivered  to  an  armature  from  any  exterior  source  is  exactly  equal 
to  the  electrical  energy  obtained  from  the  armature  plus  the  energy 
lost  in  mechanical  friction  eddy  currents  in  the  iron  and  other  small 
losses. 

Besides  producing  a  counter  torque  in  the  armature,  the  current 
tends  to  distort  or  crowd  the  lines  of  force  from  their  original  position 
in  the  magnetic  field.  This  effect  is  termed  armature  reaction.  The 
greater  part  of  the  electrical  energy  is  transmitted  to  the  external  cir- 
cuit, while  the  rest  of  the  energy,  usually  the  smaller  portion,  is  con- 
verted directly  or  indirectly  into  heat  energy  in  different  parts  of  the 
dynamo  itself. 

The  principal  armature  loss  is  that  produced  by  the  current  flowing 
against  the  internal  resistance  of  the  armature,  that  is  the  resistance 
of  the  conductors. 

The  core  loss,  is  the  energy  converted  into  heat  in  the  iron  discs  of 
the  armature  core  when  they  are  rotated  in  a  magnetic  field. 

A  small  portion  of  this  loss  is  due  to  eddy  currents  generated  in  the 
revolving  core  discs.  A  large  portion  is  due  to  magnetic  friction  which 
occurs  whenever  the  direction  of  the  lines  of  force  is  rapidly  changed 
in  a  magnetic  substance. 

This  effect  is  termed  "Hysteresis." 

The  energy  expended  by  hysteresis  is  furnished  by  the  force  which 
causes  the  change  ii  the  magnetism;  and  in  the  case  of  an  electro- 
magnet when  the  magnetism  is  reversed  by  the  magnetizing  current, 
the  energy  is  suppli-i  by  the  magnetizing  current. 

The  san<?  effect  is  produced  when  the  iron  of  the  armature  core  is 
rapidly  rotted  in  the  constant  magnetic  field. 

/This  cap  •;  differs  from  the  electro-magnet  only  in  the  fact  that  the 
magnetic  1'  nes  of  force  remain  at  rest  and  the  iron  core  is  made  to 
rotate. 


Since  the  core  is  rotated  from  the  armature  shaft,  the  energy  lost  in 
hysteresis  is  furnished  by  the  force  which  drives  the  shaft.  In  well 
designed  dynamos  the  core  loss  should  not  exceed  2%  of  its  input 
when  delivering  its  rated  output  from  the  brushes. 

The  total  per  cent  losses  in  armatures  of  constant  potential  dynamos 
varies  from  about  12%  of  the  input  to  dynamos  having  a  capacity  of 
1,000  watts  to  as  low  as  1.5%  to  2%  of  the  input  of  dynamos  of  rated 
capacity  of  about  100,000  watts  and  upwards. 

Armature  reactions  not  only  distort  the  magnetic  field,  but  also  have 
a  tendency  to  reduce  the  total  number  of  lines  of  force  from  the  magnet 
and  thereby  diminish  the  E.  M.  F.  generated  in  the  armature. 

This  effect,  however,  can  be  almost  entirely  eliminated  by  increasing 
the  strength  of  the  field;  or  in  other  words,  increasing  the  number  of 
lines  of  force  passing  through  the  core. 


Q.  75.  (1900-01.)  What  is  meant  by  the  term  separately  excited  dy- 
namos? Elucidate. 

Explain  the  term  magnifying  force  and  ampere-turns. 

Are  the  number  of  lines  of  force  directly  proportional  to  the  number 
of  ampere-turns. 

Why? 

What  metals  should  be  used  in  field  magnets  and  why? 

What  is  meant  by  magnetic  saturation? 

What  is  the  limit  for  practical  saturation  in  different  kinds  of  iron? 

What  is  the  effect  if  this  limit  is  exceeded?  In  general,  how  are  the 
field  coils  or  magnets  designed? 

Ans.  75.  A  separately  excited  dynamo  is  one  in  which  the  field 
coils  are  excited  from  some  exterior  source;  as,  for  instance,  a  voltaic 
battery,  or  another  continuous  current  dynamo. 

The  magnetizing  coils  are  wound  around  the  cores  of  the  field  mag- 
net in  such  direction  as  to  produce  a  closed  magnetic  circuit  through 
the  armature  and  has  no  connection  whatever  with  the  current  ob- 
tained from  the  brushes  by  rotating  the  armature. 

If  the  strength  of  the  exciting  current  is  not  changed,  the  difference 
of  potential  between  the  brushes  of  the  dynamo  when  the  armature  is 
rotated  at  a  uniform  speed,  remains  constant  so  long  as  the  external 
circuit  is  open;  but  when  the  external  circuit  is  closed,  the  difference 
of  potential  gradually  diminishes  as  the  strength  of  the  current  in- 
creases, owing  to  the  internal  resistance  of  the  armature  conductors 
and  the  reactions  of  the  armature  current  in  the  fields. 

The  magnetizing  force  is  that  which  produces  the  lines  of  force  in 
the  magnet.  Its  strength  is  proportional  to  the  strength  of  the  current 
flowing,  and  to  the  number  of  coils,  or  complete  turns  around  which 
the  current  circulates. 

The  total  number  of  turns  multiplied  by  the  strength  of  the  current 
in  amperes  will  give  the  magnetizing  force  in  ampere  turns. 

The  number  of  lines  of  force  produced  in  an  electro-magnet  is  not 
directly  proportional  to  the  magnetizing  force  in  ampere  turns. 

The  strength  of  the  magnet  in  lines  of  force  depends  upon  the  per- 
meability of  the  magnetic  substance  used  in  the  core. 

In  general,  wrought  iron,  soft  sheet  iron,  and  steel  have  greater 
permeability  than  cast  iron  and,  whenever  available,  should  be  used 
in  field  magnets. 

A  substance  has  reached  a  state  of  magnetic  saturation  when  it  has 
absorbed  all  the  lines  of  force  it  can  hold.  A  limit  is  never  reached 
when  actual  saturation  takes  place,  but  there  is:  a  limit  beyond  which 
it  becomes  impracticable  to  magnetize  a  substance. 

In  all  kinds  of  magnetic  substances,  the  permeability  decreases  when 
the  magnetism  is  increased  beyond  a  certain  limit.  Practical  saturation 
in  wrought  iron,  soft  sheet  iron  and  cast  steel,  is  when  there  are  120,000 

190 


and  130,000  lines  of  force  per  square  inch  of  the  iron  measured  on  a 
plane  at  right  angles  to  the  lines  of  force  in  the  magnet. 

If  these  limits  are  exceeded,  it  requires  an  enormous  increase  in  the 
ampere  turns  to  produce  a  slight  change  in  the  number  of  lines  of  force 
in  the  magnet. 

In  general,  however,  the  field  magnets  of  dynamos  are  designed  with 
the  density  of  the  lines  of  force  below  the  saturation  limits,  and  it  is 
safe  to  assume  that  any  change  in  the  strength  of  the  current  circulat- 
ing around  the  magnetizing  coils  produces  a  corresponding  change  in 
the  number  of  lines  of  force  passing  through  the  magnetic  circuit. 

Consequently,  if  the  strength  of  the  current  in  the  field  coils  of  a 
separately  excited  dynamo  is  increased  as  the  current  in  the  armature 
becomes  stronger,  the  E.  M.  F.  obtained  from  the  brushes  will  remain 
practically  constant. 

This  is  usually  accomplished  by  inserting  an  adjustable  resistance 
box  or  field  rheostat  in  the  series  with  the  battery  and  field  coils. 

Decreasing  the  resistance  as  the  load  increases  or  as  the  difference 
of  potential  between  the  brushes  tends  to  drop. 


Q.  76.  (1900-01.)  What  is  meant  by  the  term  "self-exciting"  as  ap- 
plied to  a  shunt  dynamo? 

In  well  designed  dynamos  how  is  the  resistance  of  the  field  coils  so 
proportioned  that  a  proportional  part  of  the  total  output  of  current 
will  pass  through  the  field  coils? 

Explain  the  term  "residual  magnetism." 

What  is  the  effect  in  the  building  up  of  the  E.  M.  F.? 

What  is  the  effect  upon  the  difference  of  potential  between  the 
brushes  of  a  shunt  dynamo  as  the  armature  current  becomes  stronger? 

How  is  this  effect  compensated  for? 

Ans.  76.  A  self-exciting  dynamo,  or  simply  a  shunt  dynamo,  is  a 
dynamo  in  which  the  magnetizing  of  field  coils  are  excited  from  the 
current  furnished  by  the  dynamo  itself,  the  field  coils  being  connected 
in  shunt  with  the  external  circuit  from  the  brushes. 

One  terminal  of  the  magnetizing  coil  is  connected  to  the  negative 
brush  and  the  other  to  the  binding  post  in  the  field  rheostat;  the  posi- 
tive brush  is  connected  to  the  arm  of  the  rheostat. 

If  the  resistance  of  the  rheostat  is1  cut  out  it  will  be  seen  that  the 
total  difference  of  potential  exists  between  the  terminals  and  the  mag- 
netizing coils  when  the  dynamo  is  generating  its  maximum  E.  M.  F. 

The  magnetizing  coils  of  a  shunt  dynamo,  however,  consists  of  a. 
large  number  of  turns  of  fine  copper  wire,  thus  making  the  resistance 
large  in  comparison  with  the  difference  of  potential  between  the  field 
terminals.  In  well  designed  dynamos  the  resistance  of  the  shunt  coil  is 
large  enough  to  allow  not  more  than  5  %  of  the  total  current  of  the 
dynamo  to  pass  through  the  field  coils. 

Permanent  magnetism  is  called  "Residual  Magnetism,"  since  it  re- 
sides in  the  metal  after  the  magnetizing  force  has  been  removed.  Soft 
iron  and  annealed  steel  retain  only  a  small  amount  of  magnetism. 
Chilled  iron  and  hardened  steel  retain  residual  magnetism  in  large 
quantities. 

When  a  shunt  dynamo  is  rotated  at  a  constant  speed,  an  appreciable 
length  of  time  elapses  before  the  armature  generates  a  maximum  E.  M. 
F.  after  the  field  circuit  is  closed  and  in  some  cases,  a  self-exciting 
dynamo  will  generate  no  current  until  after  it  has  been  once  separately 
excited. 

The  starting  of  a  dynamo  to  generate  an  E.  M.  F.  is  termed  the 
picking-up  or  building-up. 

If  the  field  coil  is  open  so  that  no  current  flows  through  the  mag- 
netizing coil,  the  armature  will  not  generate  any  E.  M.  F.  when  rotated; 
providing,  however,  that  the  field  magnets  were  not  permanent  mag- 
nets. 

191 


The  difference  of  potential  between  the  brushes  of  shunt  dynamos 
gradually  decreases  as  the  armature  current  increases  or  becomes 
stronger  on  account  of  the  internal  resistance  of  the  armature  con- 
ductors and  the  reaction  of  the  current  on  the  field. 

To  compensate  for  this  decrease  in  the  E.  M.  P.,  a  field  rheostat  of 
comparatively  high  resistance  is  connected  with  the  field  circuit  and  so 
adjusted  that  when  no  current  is  flowing  in  the  external  circuit,  only 
enough  current  flows  through  the  field  to  produce  the  normal  difference 
of  potential  between  the  brushes;  this  normal  difference  of  potential 
between  the  brushes  is  kept  constant,  as  the  load  increases,  by  gradually 
cutting  out  or  short-circuiting  the  resistance  coils  of  the  rheostat. 


Q.  77.     (1900-01.)     Explain  a  "series  dynamo." 

Explain  first  how  the  action  of  a  series  dynamo  differs  widely  from  a 
shunt  dynamo. 

In  the  second  place,  upon  what  does  the  difference  of  potential 
depend? 

What  comparison  in  the  coils  between  the  two  machines  can  be  made 
and  why  is  this  necessary? 

How  is  a  series  dynamo  regulated? 

How  many  methods  are  there  employed? 

How  is  this  regulation  sometimes  made  automatic? 

Can  this  regulation  be  accomplished  in  the  dynamo  itself?  If  so, 
what  is  this  style  of  a  dynamo  called? 

Ans.  77.  The  series  dynamo  is  one  where  the  magnetizing  coils  are 
connected  directly  in  series  with  the  external  circuit;  that  is,  all  the 
current  from  the  armature  circulates  around  the  magnetizing  coils  be- 
fore passing  through  the  external  circuit. 

The  action  of  a  series  dynamo  differs  widely  from  the  shunt  dyna- 
mo. In  the  first  place,  no  E.  M.  P.  is  generated  in  the  armature  unless 
the  circuit  is  closed  and  flows  from  the  brushes;  that  is,  neglecting  the 
small  E.  M.  P.  generated,  due  to  the  residual  magnetism. 

In  the  second  place,  the  difference  of  potential  between  the  brushes 
depends  upon  the  strength  of  current  flowing  from  the  armature. 

The  E.  M.  P.,  however,  is  not  directly  proportional  to  the  strength 
of  the  current  unless  the  internal  resistance  and  reactions  are  excessive. 

Compared  with  the  coils  of  a  shunt  dynamo,  the  magnetizing  coils 
of  a  series  dynamo  are  made  with  a  few  turns  of  a  large  conductor. 

This  is  necessary  because  the  coils  usually  are  required  to  carry  the 
total  current  of  the  armature;  and  the  conductor  is  made  large  enough 
to  carry  the  current  without  heating,  and  only  a  few  turns  are  neces- 
sary to  secure  the  proper  magnetizing,  since  that  is  proportional  to 
ampere  turns. 

The  E.  M.  P.  of  a  series  dynamo  is  regulated  in  three  different  ways— 
viz.: 

(1)  By  controlling  the  strength  of  the  current  in  the  external  cir- 
cuit, as  has  been  described. 

(2)  By  short-circuiting,   or  cutting  out  part  of  the  magnetizing 
coils. 

(3)  By  shunting  part  of  the  current  around  the  magnetizing  coils. 

These  methods  of  regulation  are  not,  however,  automatic;  it  is,  how- 
ever, accomplished  by  a  mechanical  movement  of  an  arm  or  contact. 
This  movement  is  sometimes  imparted  by  a  magnet  controlled  by  the 
current  from  the  armature,  but  more  often  the  E.  M.  P.  is  automatically 
regulated  in  the  dynamo  itself  by  a  combination  of  shunt  and  series 
magnetizing  cons. 

Such  dynamos  are  called  compound  or  shunt  and  series  dynamos. 

192 


Q.  78.     (1900-01.)     Describe  a  "compound  dynamo." 

What  effect  has  the  series  coil  winding  upon  the  difference  of  poten- 
tial between  the  brushes? 

What  is  this  method  of  regulating  the  difference  of  potential  between 
the  brushes  called? 

How  are  the  terminal  of  a  compound  dynamo  attached  to  the  wind- 
ings, also  to  the  outside  circuit? 

What  is  meant  by  the  term  "over-compound?" 

The  expression,  "per  cent  over-compound?" 

What  are  machines  having  one  pair  of  poles  called?  More  than  one 
pair  called?  What  are  salient  poles?  What  are  consequent  poles? 

Ans.  78.  The  compound  dynamo,  the  shunt  coils,  consist  of  a  large 
number  of  turns  of  fine  insulated  wire,  wound  upon  the  core  of  the 
magnet. 

The  series  coils  consist  of  a  few  turns  of  large  insulated  wire  and 
are  either  wound  over  the  shunt  coils  or  placed  directly  on  the  core 
itself  either  above  or  below  the  shunt  winding;  this  latter  manner  of 
winding  is  preferable  on  account  of  repairs. 

The  main  part  of  the  current  from  the  armature  flows  from  the 
positive  brush  through  the  external  circuit  thence  through  the  series 
coil  to  the  negative  brush. 

The  two  terminals  of  the  shunt  coils  are  connected  to  the  positive 
and  negative  brushes  with  a  rheostat  in  series. 

The  action  of  both  currents  of  the  series  and  shunt  coils  must  be  in 
the  same  direction  around  the  coils  to  produce  the  same  polarity  in  the 
magnet;  the  series  current  reinforcing  the  shunt  current. 

When  the  dynamo  is  not  loaded  and  the  armature  is  rotated  at  nom- 
inal speed,  the  normal  E.  M.  F.  generated  is  due  to  the  magnetic  field 
produced  by  the  shunt  coils  alone. 

Upon  closing  the  external  circuit,  however,  the  difference  of  poten- 
tial between  the  brushes  tends  to  decrease  and  would  so  continue  to 
decrease  as  in  the  shunt  machine  if  the  series  coils  were  not  called  into 
action. 

The  current  circulating  around  these,  however,  reinforces  the  mag- 
netizing force  of  the  shunt  coils  and  immediately  increases  the  lines  of 
force  in  the  field  which  in  return  raises  the  potential  between  the 
brushes  to  normal.  These  actions  are  produced  simultaneously  and, 
to  all  appearances,  the  difference  of  potential  between  the  brushes 
remains  normal  for  all  changes  of  load  in  the  external  circuit.  This 
method  of  regulating  the  E.  M.  F.  of  a  dynamo  is  called  "Compound- 
ing." 

The  terminals  of  a  dynamo  are  the  binding  posts  to  which  the  exter- 
nal circuit  is  connected;  in  a  series  or  compound  dynamo  one  terminal 
is  attached  to  the  outside  end  of  the  series  coil  and  the  other  is  con- 
nected directly  to  the  brush. 

It  is  desirable  in  a  great  many  cases  to  over-compound  a  dynamo; 
that  is,  by  adding  a  sufficient  additional  number  of  turns  in  the  series 
winding  so  as  to  increase  the  difference  of  potential  between  the  brushes 
above  the  normal  when  the  load  increases. 

This  increase  in  E.  M.  F.  is  usually  expressed  in  per  cent  over- 
compound. 

Machines  having  one  pair  of  poles  are  called  "Bi-polar"  machines. 

Machines  having  more  than  one  pair  are  called  "Multi-polar"  ma- 
chines. 

Salient  Poles. — In  all  cases  where  a  single  coil  is  used,  or  where  if 
two  coils  are  used  both  wound  in  the  same  direction,  the  poles  are 
called  salient  poles. 

Consequent  Poles. — Where  two  coils  are  used  and  wound  in  opposite 
directions  they  are  called  consequent  poles. 


Q.  79.  (1900-01.)  Into  what  three  general  types,  depending  on  the 
character  of  their  currents,  are  dynamos  divided?  Describe  them. 

In  constant  potential  dynamos  and  generators,  describe  the  windings 
of  the  pole  pieces  or  field  core  and  the  direction  of  the  lines  of  force,  etc. 

Why  are  four  brushes  or  more  needed  on  a  multi-polar  machine? 

How  is  the  regulation  of  multi-polar  machines  accomplished? 

Ans.  79.  Dynamos  are  divided  into  three  general  types,  depending 
upon  the  character  of  their  currents. 

(1)  "Constant  potential  dynamos,"  in  which  the  B.  M.  F.  remains 
constant  and  the  strength  of  the  current   (continuous)   changes  with 
the  load. 

(2)  "Constant  current  dynamos,"  in  which  the  strength  of  the  cur- 
rent  {continuous  or  pulsating)    remains  constant  and   the   E.   M.   F. 
changes  with  the  load. 

(3)  "Alternating  current  dynamos,"  the  current  from  which  alter- 
nates or  reverses  direction  with  great  rapidity. 

The  previous  questions  and  answers  have  demonstrated  the  principles 
and  regulations  of  only  one  form  of  constant  potential  dynamos  that, 
in  which  an  armature  was  rotated  between  only  one  pair  of  poles.  The- 
oretically, however,  constant  potential  dynamos  can  be  built  with  an 
armature  revolving  between  any  number  of  pairs  of  poles.  Such  ma- 
chines are  termed  multi-polar  dynamos. 

In  multi-polar  dynamos,  the  pole-pieces  and  field  cores  are  fastened 
into  one  magnetic  yoke  more  or  less  circular  in  shape. 

A  magnetizing  coil  is  wound  upon  each  field  core  and  the  four  coils 
are  connected  in  series  in  such  a  manner  that  when  the  current  circu- 
lates around  the  coil  it  produces  first  a  north  pole  and  then  a  south  pole. 

The  lines  of  force  from  each  field  core  divide  into  two  magnetic  cir- 
cuits in  the  yoke  and  armature. 

Their  density  is  practically  uniform,  however,  when  they  pass  from 
the  north  pole  into  the  armature  core,  or  from  the  armature  core  into 
the  south  pole. 

In  nearly  all  multi-polar  dynamos  this  same  principle  of  polarity  is 
applied;  that  is,  every  other  pole  piece  is  of  like  polarity  and  lines  of 
force  from  each  core  divide  into  the  magnetic  circuit,  in  the  armature 
and  in  the  field  yoke. 

The  process  of  generating  and  B.  M.  F.  is  similar  to  that  in  a  bi- 
polar machine,  but  there  are  some  points  which  should  be  understood. 

In  the  case  of  a  four-pole,  ring-core  closed-winding  machine,  by 
tracing  out  the  direction  in  which  the  E.  M.  F.  tends  to  act  upon  the 
conductors,  it  will  be  seen  that  there  are  four  points  where  the  E.  M. 
F.  tends  to  act  in  opposite  directions.  The  action  of  the  electromotive 
forces  is  to  meet  between  one  pair  of  poles  and  divide  between  the  next 
pair,  etc.,  and  the  segments  at  the  two  points  of  meeting  have  the  same 
potential  and  form  two  positive  neutral  points  of  the  commutator,  while 
those  segments  at  the  point  where  the  current  divides  form  the  two 
neutral  points  of  the  commutator.  Hence  the  necessity  of  four  brushes, 
two  positive  and  two  negative.  The  two  positive  brushes  being  con- 
nected in  parallel  to  one  terminal  of  the  external  circuit  and  the  two 
negative  brushes  are  connected  to  the  other  terminal  of  the  external 
ci'ffcuit. 

It  is  possible,  however,  to  connect  and  group  the  conductors  in  an 
armature  for  a  multi-polar  dynamo  so  that  the  current  will  divide  into 
two  circuits  only,  this  winding  is  termed  a  series  winding. 

The  .regulation  of  multi-polar  dynamos  for  constant  potential  is  ac- 
complished by  changing  of  the  strength  of  the  magnetizing  force  as  in 
the  bi-polar  machines. 

core  and  all  connected  together  in  parallel,  or  series,  as  is  most  con- 
lenient, 

194 


Q.  80.  (1900-01.)  What  is  meant  by  the  efficiency  of  constant  po- 
tential dynamos? 

In  a  compound  dynamo,  the  series  coils  are  wound  on  each  field 

What  is  the  mechanical  energy  delivered  to  the  armature  shaft 
called? 

To  what  is  it  equal? 

What  is  the  electrical  energy  appearing  in  the  external  circuit  called? 

To  what  is  it  equal? 

What  is  the  energy  converted  into  heat,  directly  or  indirectly,  called? 

How  would  you  flnd  the  per  cent  efficiency  of  a  dynamo? 

How  would  you  find  the  total  per  cent  loss  of  a  dynamo? 

Upon  what  does  the  efficiency  of  a  dynamo  depend? 

How  would  you  find  the  input  necessary  to  drive  a  dynamo  when  its 
output  and  efficiency  at  that  output  are  given? 

How  would  you  find  the  output  of  dynamo  when  its  input  and  its 
efficiency  at  that  input  are  given? 

Name  the  four  classes  into  which  the  total  loss  of  power  in  a  dyna- 
mo, ordinarily  is  due  to. 

Describe  each,  briefly. 

Ans.  80.  If  in  converting  or  transforming  a  certain  amount  of  me- 
chanical energy  into  electrical  energy,  the  energy  which  is  manifested 
in  une  form  disappears,  and  the  same  quantity  appears  in  another 
form,  or  in  several  different  forms,  then  the  amount  of  energy  delivered 
at  the  armature  shaft  is  always  equal  to  the  energy  appearing  in  the 
external  circuit,  plus  the  energy  converted  into  heat  in  the  dynamo 
itself. 

The  mechanical  energy  delivered  to  the  armature  shaft  is  termed 
the  input. 

The  input  is  always  equal  to  the  output  at  the  brushes  plus  the 
losses  in  the  machine  itself. 

The  electrical  energy  appearing  in  the  external  circuit  from  the 
brushes  is  termed  the  output. 

The  output  plus  the  energy  lost  in  the  dynamo  itself  is  equal  to  the 
input. 

Energy  lost  or  converted  into  heat  is  termed  energy  losses  or  simply 
losses. 

To  find  the  per  cent  efficiency  of  a  dynamo: 

Rule. — Multiply  the  output  in  watts  by  100  and  divide  the  product  by 
the  input  in  watts. 

To  find  the  total  per  cent  loss  in  a  dynamo : 

Rule. — Multiply  the  difference  in  watts  between  the  output  and  the 
input  by  100  and  divide  this  product  by  the  input  in  watts. 

The  efficiency  of  a  dynamo  depends  upon  its  character,  construction, 
condition  when  tested,  its  capacity,  losses  and  various  conditions;  in 
fact,  two  dynamos  of  the  same  construction  and  capacity  seldom  show 
exactly  the  same  efficiencies. 

To  find  the  input  necessary  to  drive  a  dynamo  when  its!  output  and 
its  efficiency  at  that  output  is  known: 

Rule. — Divide  the  output  in  watts  by  the  per  cent  efficiency  and 
multiply  the  quotient  by  100. 

To  find  the  output  of  a  dynamo  when  its  efficiency  is  known  at  that 
input: 

Rule.— Multiply  the  input  by  the  per  cent  efficiency  and  divide  this 
product  by  100. 

(1)  Mechanical  friction  loss. 

(2)  Core  loss. 

(3)  Field  loss. 

(4)  Armature  loss. 

Friction  loss,  due  to  mechanical  friction  which  takes  place  between 
tne  bearings  and  journals.  The  brushes  rubbing  on  the  commutator 
produce  some  friction,  and  consequent  loss.  Under  ordinary  condition^ 

195 


the  loss  in  mechanical  friction  should  not  exceed  5%  of  the  input  of  the 
dynamo. 

Core  losses  is  the  energy  converted  into  heat  in  the  iron  discs  when 
they,  are  rotated  in  the  magnetic  field.  A  small  portion  of  this  loss 
is  due  to  eddy  currents  generated  in  the  revolving  core.  The  larger 
portion  of  this  loss  is  due  to  a  magnetic  friction  which  occurs  whenever 
the  direction  of  the  lines  of  force  is  rapidly  changed  in  a  magnetic 
substance.  The  effect  produced  by  this  rapid  change  in  direction  of  the 
magnetizing  current,  the  iron  or  steel  of  the  core  becomes  heated  which 
necessitates  a  certain  amount  of  expenditure  of  energy.  This  effect  is 
called  hysteresis. 

In  a  well  designed  dynamo  the  core  loss  should  not  exceed  2%  of  its 
input  when  delivering  its  rated  output  from  the  brushes. 

Field  Losses. — In  self-exciting  dynamos  a  portion  of  the  electrical 
energy  generated  in  the  armature  is  required  for  the  excitation  of  the 
field  magnets. 

This  energy  is  considered  as  a  loss,  since  it  does  not  appear  in  the 
external  circuit  and  is  entirely  dissipated  in  the  form  of  heat. 

The  per  cent  loss  in  the  field  coils  of  dynamos  varies  from  10%  of 
the  input  to  dynamos  having  an  output  of  1,000  watts  to  as  low  as  1.5% 
to  2%  of  the  input  of  dynamos  having  an  output  100,000  watts  and  up- 
wards. 

The  armature  loss  proper  is  usually  termed  the  copper  or  wire  loss, 
and  is  produced  by  the  current  flowing  against  the  internal  resistance 
GJ.  the  armature. 

The  per  cent  loss  in  armatures  of  constant  potential  dynamos  varies 
from  12%  of  the  input  of  dynamos  having  a  rated  capacity  of  about 
1,000  watts,  to  as  low  as  1.5%  to  2%  of  the  input  of  dynamos  having 
the  rated  capacity  of  about  100,000  watts  and  upwards. 


196 


Technical,   Mathematics,   Science, 
Mechanics,  Physics,  Etc. 


Q.  10.   (1896-7.)     How  to  find  velocity  of  water  in  a  pipe,  knowing 
the  pressure? 

Ans.  10.    Formula:  — 


Vm=2.3l5 


/hd 
\fl.+. 


I25d 


in  which  Vm  =  mean  velocity, 
h  =  total  height  in  feet. 
1  =  length  of  pipe  (straight), 
d  =  diameter  of  pipe, 
f  —  coefficient  of  friction. 

The  pressure  being  given  it  is  reduced  to  head  by  dividing  the 
pressure  by  .434. 
Assume: 

h  =  300  ft. 

1  =  100  ft. 

d  =  6". 

f  =.0214. 
Then :     substituting 


Vm=2.315. 


.0214  X  100+  .75 

I 
=2.315, 


=  2.315  X  24.9  =  57.6435  ft.  per  second. 

[The  constant  2.315  is  the  product  of  several  factors  that  enter  into 
the  calculation  and  are  calculated  once  for  all  in  forming  the  constant. 
These  factors  are: 

gxk          32.X.862 

= =  2.315 

1/144  12 

g  =  gravity,  32.2. 

k  =  vena  contracta,  the  reduction  in  the  diameter  of  the  stream  of 
water  after  it  leaves  the  end  of  the  pipe,  or  nozzle. 

12  =  square  root  of  144,  because  some  of  the  factors  in  this  equation 
are  given  in  feet  and  the  square  root  of  144  shows  the  relation  between 
"inches  diameter"  and  "feet  diameter,"  giving  the  proportionate  area.] 


Q.  15.  (1896-7.)     How  to  find  the  area  of  a  segment? 

Ans.  15.  Divide  the  diameter  of  the  circle  by  the  height  of  the 
segment,  subtract  .608  from  the  quotient  and  extract  the  square  root 
of  the  remainder.  This  result  multiplied  by  four  times  the  square  of 

197 


the  height  of  the  segment  and  divided  by  three  will  give  the  area,  near- 
ly.    Ex.: 


height  of  segment, 
diameter  of  circle. 


Q.  19.  (1896-7.)     What  is  momentum? 
Ans.  19.     Product  of  mass  X  velocity. 


Q.  24.  (1896-7.)  Can  water  be  evaporated  with  exhaust  steam  of 
atmospheric  pressure,  and  if  so,  how? 

Ans.  24.  Yes;  provided  the  pressure  upon  the  water  be  kept  below 
that  of  the  atmosphere;  in  other  words,  under  a  vacuum.  The  lower 
the  pressure  or  the  greater  the  vacuum,  the  more  rapid  will  be  the 
evaporation. 


Q.  26.  (1896-7.)  Is  steam  at  thirty  Ibs.  pressure  absolute  or  air  at 
sea  level,  the  heavier? 

Ans.  26.  Little  difference;  air  being  13.141  cubic  feet  to  the  pound, 
and  steam  13.480  cubic  feet  to  the  pound,  assuming  the  air  to  be  dry 
and  at  the  standard  temperature,  62°  F. 


Q.  27.  (1896-7.)  What  is  the  expansion  of  a  wrought  iron  pipe  75 
ft.  long,  erected  at  an  external  temperature  of  20°  F.,  and  then  charged 
with  steam  at  100  Ibs.  gage  pressure? 

Ans.  27.  Temperature  of  steam  100  Ibs.  pressure  is  337.7°.  Co-ef- 
ficient of  expansion  for  wrought  iron  is  .0000067302.  Then  337.7  —  20 
=  317.7,  and  75  X  12  X  317.7  X  .0000067302  =  1.924". 


Q.  28.  (1896-7.)  How  determine  the  pressure  per  square  inch  at  the 
base  of  a  water  column  125  ft.  high? 

Ans.  28.  Generally  speaking,  the  pressure  in  pounds  per  square 
inch  equals  head  in  feet  X  .434.  To  be  accurate,  however,  the  tempera- 
ture of  the  water  should  be  taken  into  account  and  the  constant  for  such 
will  be: 

Wgt.  of  cu.  ft.  water  at  obsv'd  temp. 


144 
Then,  125  ft.  X  .434  =  54.25  Ibs. 


Q.  33.  (1896-7.)  How  many  foot  pounds  of  work  required  to  change 
one  pound  of  water  at  32°  F.,  into  steam  at  212°? 

Ans.  33.  The  sensible  heat  to  raise  one  pound  of  water  from  32° 
to  212°  is  180.9  heat  units.  The  latent  heat  of  evaporation  at  212°  is 
965.7  heat  units  and  the  sum  of  the  two,  or  total  heat,  is  1146.6  heat 
units.  Since  one  heat  unit  equals  778  foot-pounds  of  work,  there  will 
be  required  to  change  one  Ib.  of  water  at  32°  into  steam  at  212°,  1146.6 
X  778,  or  892054.8  foot-pounds  of  work. 

198 


Q.  35.  (1896-7.)  Name  the  five  most  famous  modern  steam  en- 
gineers. 

Ans.  35.  Watt,  Stephenson,  Evans,  Rankine,  Corliss  (probably 
Newcomen,  Fulton,  Pairbairn  and  Ericcson,  or  some  names  from  non- 
English  speaking  countries  should  be  included).  I  have  purposely 
avoided  naming  any  men  living  at  this  day. 


Q.  36.   (1896-7.)     What  is  adiabatic  expansion? 

Ans.  36.     Adiabatic  expansion  is  the  curve  formed  by  a  perfect  gas 
without  loss  of  heat  or  transmission  to  or  from  the  same. 


Q.  45.  (1896-7.)  Will  the  pressure  per  sq.  in.  on  two  columns  of 
the  same  liquid,  of  equal  height  but  different  diameters,  be  the  same? 

Ans.  45.  The  pressure  per  square  inch  of  a  column  of  liquid  de- 
pends on  its  height  and  temperature,  and  is  not  affected  by  its  diameter, 
and  therefore  the  two  columns,  as  contemplated  in  the  problem,  have 
equal  pressures.  If  they  stand  close  together  this  is  so,  but  if  one  be 
at  the  equator  and  the  other  at  the  pole  of  the  earth  there  will  be  a 
difference,  owing  to  the  effect  of  gravity.  The  one  having  the  greater 
sensible  heat,  and  therefore  the  lesser  specific  weight,  will  show  the 
lesser  pressure.  If  one  be  very  much  smaller  than  the  other,  cohesion 
of  the  retaining  walls  and  attraction  for  same  may  lessen  the  pressure 
of  the  smaller  column.  This  is  more  particularly  true  when  the  diam- 
eter of  the  column  is  very  small,  say  the  fractional  part  of  an  inch. 
Unusually  thick  and  dense  retaining  walls'  about  the  column  may  also 
act  as  a  cliff  or  wall  does  upon  a  plumb-bob,  and,  by  exerting  a  side 
pull,  lessen  the  downward  effect  of  the  column  of  liquid.  Foreign  par- 
ticles in  suspension  or  solution  in  the  liquid  may  reduce  or  increase  its 
weight  and  pressure. 

Q.  49.  (1896-7.)  What  abundant  metal  of  commerce  has  the  high- 
est thermal  conductivity? 

Ans.  49.  Copper,  which  excels  all  other  metals  but  silver  in  conduc- 
tivity of  heat,  and  because  silver  is  not  properly  an  abundant  metal. 


Q.  53.     (1896-7.)     What  is  work? 

Ans.  53.  Work  is  the  overcoming  of  resistance  through  space  or 
through  a  certain  distance.  It  is  measured  by  the  product  of  the  resis- 
tance into  the  space  through  which  it  is  overcome.  It  is  also  measured 
by  the  product  of  the  moving  force  into  the  distance  through  which 
the  force  acts  in  overcoming  the  resistance.  Thus,  in  lifting  a  body 
from  the  earth  against  the  attraction  of  gravity,  the  resistance  is  the 
weight  of  the  body  and  the  product  of  this  weight  into  the  height  the 
body  is  lifted  is  the  work  done. 


Q.  56.     (1896-7.)     What  is  heat? 

Ans.  56.  Heat  is  not  a  substance,  but  is  a  form  of  energy.  Heat  is 
treated  in  scientific  books  under  the  heading  "Thermo-Dynamics,"  from 
two  Greek  words,  signifying  heat  power.  The  word  "heat"  is  commonly 
used  in  two  senses — -first,  to  express  the  sensation  of  warmth;  second, 
the  state  of  things  in  a  body  that  causes  that  sensation. 

Heat,  as  a  form  of  energy,  is  subject  to  the  general  laws  governing 
every  form  of  energy  and  controlling  all  matter  in  motion,  whether 
that  be  molecular,  or  movement  of  the  mass.  Heat  is  not  a  substance. 
It  can  be  absorbed  and  reflected  the  same  as  a  ray  of  light.  It  can  be 
conducted  through  a  substance  or  between  two  bodies  in  contact.  When- 

199 


ever  there  exists  a  difference  in  temperature  between  two  bodies,  and 
particularly  if  in  proximity  to  each  other,  there  is  a  tendency  to  an 
exchange  of  heat  and  an  equalization  of  temperature. 

Heat,  like  other  energy,  cannot  be  destroyed,  but  can  be  altered  in 
form. 

Heat  usefully  employed  plus  that  lost  or  wasted  always  equals  in 
amount  the  total  heat  applied. 

Heat  and  temperature  must  not  be  confused  with  each  other.  A 
dipperful  of  water  and  a  pailful  may  be  of  the  same  temperature  and 
yet  contain  very  different  total  amounts  of  heat.  Quantity  of  heat  in  a 
body  depends  upon  the  mass,  temperature  and  specific  heat  of  the  sub- 
stance. Heat  can  be  produced  or  brought  into  evidence  by  physical  or 
chemical  forces,  and  may  be  a  measure  thereof — being  always  propor- 
tional to  the  effort  expended. 


Q.  58.  (1896-7.)  What  is  the  latent  heat  of  steam?  Explain  how  it 
may  be  found. 

Ans.  58.  Latent  heat  of  steam  is  the  quantity  of  heat  which  disap- 
pears or  becomes  concealed  in  it  while  producing  some  change  in  it 
other  than  a  rise  in  temperature.  By  reversing  this  change,  the  quan- 
tity of  heat  which  had  disappeared  will  be  reproduced.  It  may  be 
observed  that  heat  disappears  when  water  passes  from  the  liquid  to  the 
gaseous  state.  At  212°  temperature  this  has  been  determined  by  experi- 
ment to  be  nearly  966  units  of  heat.  From  212°  to  32°  there  are  180 
heat  units  absorbed;  then  966  -f- 180  =  5.37  times  as  much  heat  needed 
to  convert  water  into  steam  as  is  required  to  raise  its.  temperature  to 
the  boiling  point.  To  find  latent  heat  of  steam,  multiply  the  sensible 
heat  by  .3,  add  1115°,  and  we  get  the  total  heat,  from  which  we  sub- 
tract the  sensible  heat  to  find  the  latent  heat. 


Q.  66.  (1896-7.)  At  what  temperature  are  carbonates  and  sulphates 
of  lime  practically  all  deposited  from  feed  waters? 

Ans.  66.  Carbonates  of  lime  are  practically  all  deposited  from  feed 
water  at  212°  F.;  sulphate  of  lime  at  about  280°  F.  to  300°  F. 


Q.  67.     (1896-7.)     What  is  meant  by  sublimation? 

Ans.  67.  To  use  the  term  literally  and  chemically  sublimation  is  an 
operation  by  which  a  solid  body  is  changed  by  heat  into  vapor  and 
then  condensed  into  a  solid  form  again,  without  changing  into  a  liquid. 
To  use  it  figuratively,  it  is  the  act  of  heightening,  refining  and  exalting 
that  which  is  highly  refined,  purified  or  improved.  Sublimate  (chemi- 
cally)— the  result  of  the  process  of  sublimation — a  body  obtained  in 
the  solid  state  by  the  cooling  of  its  vapor.  For  example,  iodine,  sal-am- 
moniac, mercuric  chloride  (corrosive  sublimate),  camphor,  etc. 


Q.  70.     (1896-7.)     What  is  isothermal  expansion? 

Ans.  70.  Isothermal  expansion  is  the  application  of  what  is  known 
as  Mariotte's  law  of  the  expansion  of  gases — that  the  volume  of  all 
elastic  gases  and  vapors  is  inversely  as  the  pressure.  Thus,  if  a  given 
volume  be  allowed  to  expand  to  two  such  volumes,  the  pressure  will  be 
reduced  one-half;  if  to  three  volumes,  to  one-third,  and  so  on.  If  com- 
pressed, the  pressure  will  rise  in  proportion  to  the  reduction  of  volume. 

This  law  is  found  to  be  correct  only  when  expansion  takes  place  at 
constant  temperature,  and  is  therefore  called  "isothermal  (equal  tem- 
perature) expansion." 


Q.  74.  (1896-7.)  What  is  the  final  temperature  of  a  mixture  of  3 
Ibs  o'f  ice  at  10°  F.,  with  20  Ibs.  of  water  at  60°,  there  being  no  loss  of 
heat? 

Ans.  74.  The  specific  heat  of  ice  is:  .504.  Heat  of  liquefaction  of 
ice  is  142.65.  Heat  required  to  bring  1  Ib.  of  ice  from  10°  to  32°  = 
32  — 10  =  22  units.  22  X  3  X  .504  =  33.264  heat  units. 

Heat  units  required  to  convert  ice  at  32°  into  water  at  32°  is  142.65 
X  3  =  427.95. 

Total  heat  needed  to  convert  3  Ibs.  ice  at  10°  into  water  at  32°  = 
427.95  +  33.264  =461.214  units. 

Total  heat  in  20  Ibs.  water  at  60°  =  20  X  60  =  1200. 

Total  heat  in  3  Ibs.  water  at  32°  =  3  X  32  =  96. 

Total,  1296  heat  units. 

The  total  sensible  heat  in  the  mixture  would  be  1296—461.214  = 
834.768  units. 

834.768  -j-  23    (20  +  3)  =  36.295°,  temperature  of  the  mixture. 


Q.  75.  (1896-7.)  What  is  meant  by  the  external  and  internal  latent 
heat  of  steam? 

Ans.  75.     The  total  heat  in  steam  includes  three  elements: 

1.  Heat  required  to  raise  the  temperature  of  the  water  to  the  tem- 
perature of  the  steam. 

2.  Heat 'required  to  evaporate  the  water  at  that  temperature  called 
the  internal  latent  heat. 

3.  The  latent  heat  of  volume,  or  the  external  work  done  by  the 
steam  in  making  room  for  itself  against  the  pressure  of  the  superin- 
cumbent atmosphere. 

The  sum  of  the  last  two  taken  together  is  called  the  latent  heat  of 
steam.  The  heat  required  to  generate  1  Ib.  of  steam  at  212°  from 
water  at  32°  is: 

Sensible  heat  to  raise  water  from  32°  to  212°=  180.9° 

Latent  heat:    (1)   of  the  formation  of  steam 

at  212°  =894.0° 

(2)  of  expansion  against  atmospheric  pres- 
ure,  2116.4  Ibs.  per  sq.  ft.  X  26.36  cu.  ft. 
=  55786  ft.  lbs.H-778  ..  ..=  71.7°  965.7° 


Total  above  32° 1146.6° 


Q.  77.  (1896-7.)  What  is  the  absolute  zero  of  temperature,  and  how 
found? 

Ans.  77.  The  absolute  zero  of  temperature  is  a  condition  in  which 
there  is  supposed  to  be  no  heat  whatever,  and  is  taken  to  be  461°  below 
zero  of  the  Fahrenheit  scale.  According  to  the  law  of  Gay  Lussac,  a 
perfect  gas  will  decrease  1/461  part  of  its  volume  at  32°  F.  for  each 
degree  of  heat  abstracted. 

Thus,  if  from  a  given  volume  of  gas  at  zero  1°  of  heat  be  taken,  the 
resulting  volume  will  be  460/461  of  the  former  volume,  and  if  2°  he 
taken  away  the  resulting  volume  will  be  459/461  of  the  original,  and  so 
on  down  until  the  temperature  has  reached  461°  below  zero,  when,  in 
theory,  gas  will  have  no  volume  and  no  heat.  It  is  not  possible  to  prove 
to  a  certainty  the  truth  of  this  law,  because  we  have  no  means  of  pro- 
ducing so  low  a  temperature  nor  any  means  of  measuring  the  quantity 
of  heat. 

Experiments  have  been  carried  so  far  in  this  line  as  to  remove  all 
reasonable  doubt  of  the  practical  application  of  the  law.  Regnault'S 
and  Tate's  experiments  place  the  absolute  zero  at  458.71°  F.  below  zero. 


Q.  78.     (1896-7.)     What  is  meant  by  the  British  Thermal  unit? 

Ans.  78.  The  British  unit  of  heat,  or  the  British  thermal  unit,  or 
B.T.U.,  is  that  quantity  of  heat  which  is  required  to  raise  the  tempera- 
ture of  1  Ib.  of  pure  water  1°  F.  at  or  near  39.1°  F.,  the  temperature  of 
its  maximum  density.  Peabody's  definition:  The  heat  required  to  raise 
1  Ib.  water  from  62°  to  63°  F.  is  not  generally  accepted.  The  mechani- 
cal equivalent  of  heat,  or  of  a  unit  of  heat,  is  778  foot-lbs.  of  energy. 


Q.  79.     (1896-7.)     What  is  meant  by  specific  heat? 

Ans.  79.  Specific  heat  is  a  term  used  to  define  thermal  capacity,  and 
is  usually  a  fractional  number  or  coefficient  which  expresses  the  rela- 
tive heat  storing  capacities  of  different  substances  as  compared  with 
some  substance  taken  as  a  basis  and  which  has  an  assigned  value  of  1. 

The  unit  or  basis  taken  for  specific  heat  tables  in  which  solids  and 
liquids  figure  is  usually  water  and  the  quantities  of  heat  taken  up  by 
equal  weights  of  various  substances,  for  equal  increase  in  temperature, 
are  compared  with  a  like  weight  of  water. 

Illustration:  Thus,  if  the  base  of  the  table  is  water,  with  an 
assigned  value  of  1,  we  find  the  specific  heat  of  wrought  iron  is  .1138, 
this  fraction  representing  the  relative  quantity  of  heat  which  1  Ib.  of 
wrought  iron  will  absorb  or  give  out,  as  compared  with  the  same  weight 
of  water,  the  temperatures  of  both  being  raised  or  lowered  equally. 


Q.  82.  (18y6-7.)  Can  steam  be  liquefied  by  pressure -without  the 
radiation  of  heat? 

^ns.  82.  No.  As  no  heat  is  allowed  to  escape  the  steam  is  com- 
presse,  adiabatically,  work  done  on  the  steam  by  the  act  of  compress- 
ing it,  which  represents  so  much  added  heat,  enabling  it  to  exist  at 
a  higher  pressure.  The  more  it  is  compressed  the  greater  will  be  the 
increase  of  temperature.  Not  only  is  the  heat  sufficient  to  maintain  the 
steam  in  its  gaseous  form,  but  if  there  is  water  in  the  cylinder  at  same 
temperature  the  heat  generated  by  the  compression  is  sufficient  to  turn 
all  or  part  of  it  into  steam. 


Q.  83.  (1896-7.)  Is  oil  or  grease  the  better  general  lubricant  for 
bearings? 

Ans.  83.  The  elements  which  determine  the  value  of  a  lubricant  are 
the  cost  due  to  consumption  of  the  lubricant,  the  expense  for  fuel  used 
in  overcoming  frictional  resistance  during  use  of  lubricant  and  the 
cost  of  renewing  worn  journals  and  boxes. 

The  economy  of  one  oil  over  another,  so  far  as  durability  is  con- 
cerned, is  simply  proportional  to  the  rate  at  which  it  can  insinuate  itself 
into  and  flow  out  of  the  bearing.  This  ability  is  governed  by  the  vis- 
cosity of  the  lubricant  and  the  amount  of  foreign  matter  in  it. 

Where  the  lubricating  film  of  oil  or  other  substance  is  thick  and 
large  amounts  will  pass  through  a  box  or  bearing,  the  greater  the  vis- 
cosity the  less  will  be  the  flow  of  waste.  Other  things  being  equal,  an 
oil  with  imperfect  fluidity  will  be  the  cheapest.  When  the  feed  of  an 
oil  is  restricted,  as  in  case  of  crank  pins,  and  yet  the  flow  must  be 
constant  and  steady  at  normal  temperature,  it  is  hardly  practicable 
to  feed  greases  or  heavy  oils.  ' 

In  a  general  way,  for  heavy  pressure  and  slow  speed,  grease  or 
heavy  oil  is  best.  For  light  pressure  or  high  speeds  oil  is  best.  For 
ordinary  factory  shafting,  grease  is  clean,  convenient  and  cheap. 


Q.  85.     (1896-7.)     What  is  meant  by  the  critical  temperature? 
Ans.  85.     There  appears  to  exist  for  each  and  every  gas  a  tempera- 
ture above  which  it  cannot  be  liquefied  at  any  pressure.    This  is  called 

202 


the  critical  temperature — ail  gases  or  vapors  can  be  liquefied  below  that 
temperature  if  sufficient  pressure  is  used. 


Q.  97.  (1896-7.)  How  large  a  steam  radiator  needed  to  heat  an 
office  15  ft.  by  20  ft.,  with  a  12  ft.  ceiling? 

Ans.  97.  The  size  of  steam  radiator  to  heat  an  office  depends  upon 
location,  surroundings,  exposure,  cubical  contents  of  the  office,  number 
of  doors  and  windows,  external  temperature,  style  of  heater,  whether 
direct  or  indirect,  etc.,  etc. 

Assuming  that  the  office  is  favorably  situated  as  one  of  a  suite  in  a 
well  constructed  building  with  but  one  out-door  wall  and  window  and 
located  in  the  northern  temperate  zone  of  the  U.S.A.,  then  48  sq.  ft. 
of  radiation  will  suffice  to  warm  the  room  to  70°  in  zero  weather.  To 
find  it,  proceed  as  follows:  Get  cubical  contents  of  room,  then  15'  X  20' 
X  12'  =  3,600  cu.  ft.  space;  divide  by  factor  75,  gives  48  sq.  ft.  for  an 
answer,  calling  for  144  lineal  feet  of  1"  pipe,  or  125  lineal  feet  of  IW 
pipe,  or  an  ordinary  12-loop  sectional  radiator  containing  4  ft.  to  the 
loop.  Experience  and  conditions  determine  the  factor  to  be  used,  which 
varies  from  40  or  50  to  200  or  250,  to  meet  the  conditions  mentioned  in 
introduction  of  this  answer. 


Q.  22.  (1897-8.)  What  is  a  calorimeter,  how  constructed,  and  for 
what  purpose  used? 

Ans.  22.  The  calorimeter  is,  as  its  name  implies,  an  apparatus  for 
measuring  heat.  In  stationary  engineering  there  are  two  kinds  of 
calorimeters,  one  for  the  purpose  of  ascertaining  the  amount  of  mois- 
ture in  steam,  the  other  for  ascertaining  the  heat  value  of  coal. 

A  type  of  the  first  kind  is  the  barrel  calorimeter.  This  consists 
essentially  of  a  barrel  of  water  with  a  steam  pipe  leading  into  it.  The 
steam  to  be  tested  is  allowed  to  run  into  and  be  condensed  in  the 
water.  A  comparison  of  the  increase  in  weight  and  in  temperature  of 
the  water  will  show  the  percentage  of  water  carried  in  with  the  steam. 

The  second  form  consists  essentially  of  one  vessel  placed  within 
another.  The  outer  vessel  contains  water.  A  small  portion  of  the  coal 
to  be  tested  is  placed  in  the  inner  vessel,  which  is  supplied  with  oxygen 
gas.  The  coal  is  rapidly  and  completely  burned  in  this  gas,  and  the 
amount  of  heat  generated  by  its  combustion  is  measured  by  the  rise 
in  temperature  of  the  water. 


Q.  26.  (1897-8.)  What  is  meant  by  the  "viscosity"  of  oil?  How  can 
the  relative  viscosities  of  two  specimens  be  conveniently  determined  and 
to  what  extent  is  it  a  criterion  of  the  value  of  the  oil? 

Ans.  26.  Viscosity  is  the  friction  of  the  particles  upon  each  other. 
The  relative  viscosity  of  two  specimens  of  oil  may  be  approximately 
determined  by  filling  a  small  vessel  with  the  oils,  one  after  the  other, 
allowing  them  to  run  out  through  small  orifices  and  noting  the  time 
required  to  discharge  equal  quantities.  The  specimen  taking  the  longer 
time  to  run  out  is  proportionately  more  viscous. 

The  viscosity  is  often  taken  as  an  index  of  the  value  of  the  oil  for 
heavy  work. 


Q.  27.  (1897-8.)  What  is  the  rule  for  finding  the  number  of  foot- 
pounds of  work  necessary  to  change  the  velocity  of  a  body  of  known 
weight  from  one  value  to  another? 

Ans.  27.     Subtract  the  square  of  the  lesser  velocity  from  the  square 
"203 


of  the  greater  velocity  and  multiply  the  remainder  by  the  weight.  Di- 
vide that  product  by  64.4  and  the  result  will  be  the  required  number  of 
foot-pounds.  Velocities  are  taken  in  feet  per  second,  and  weight  in 
pounds: 

This  is  expressed  algebraically  as  follows: 
W.(V2— Vi2) 


64.4 

in  which  W  =  weight  in  pounds,  Vt  the  less  and  V  the  greater  velocity 
in  feet  per  second,  E=  energy  in  foot  pounds  (work). 

Q.  31.  (1897-8.)  What  is  the  rule  for  obtaining  the  centrifugal 
force  of  a  given  weight,  moving  in  a  circle  of  a  given  radius,  with  a 
given  velocity? 

Ans.  31.    The  formula  for  centrifugal  force  is 
WV2 


32. 2R 
in  which  R  is  expressed  in  feet  and  fractions  thereof. 

Multiply  the  weight  by  the  square  of  the  velocity  and  divide  the 
product  by  32.2  times  the  radius.  The  radius  is  to  be  taken  in  feet, 
velocities  in  feet-per-second,  and  weight  in  pounds. 


Q.  32.     (1897-8.)     For  the  engine  of  Q.  28  draw  a  diagram  to  scale,  by 
help  of  the  rule  of  Ans.  31,  such  that  horizontal  distances  shall  repre- 
sent piston-positions,  and  vertical  distances  the  force  exerted  by  the 
inertia  of  the  piston,  piston-rod,  and  cross-head,  at  each  piston-position. 
Ans.  32.    Lay  off  as  before  a  horizontal  line  a,  b,  Fig.  3,  to  represent 


the  stroke.  Imagine  that  the  total  weight  of  the  parts  is  upon  the 
crank-pin  and  calculate  what  the  centrifugal  force  would  be.  Lay  off  a 
line  a,  c,  proportional  to  this  force,  vertically  downward  from  the  end 
of  the  line  a,  b,  which  represents  the  commencement  of  the  stroke. 
Lay  off  a  line  b,  d,  vertically  upward  from  the  other  end  of  the  line  a,  b, 
and  connect  the  points  c,  and  d,  by  a  straight  line.  Then  will  points  in 
the  line  a,  b,  represent  piston  positions  and  the  vertical  distances  from 
said  points  to  the  line  c,  d,  will  represent  the  force  of  inertia  at  the 
corresponding  points  in  the  stroke.  Distances  downward  from  the  line 
a,  b,  represent  pulls  on  the  connecting  rod;  distances  upward,  pressures 
upon  the  same.  The  above  diagram  neglects  the  effect  of  the  change  in 
angularity  of 'the  connecting  rod.  This  may  easily  be  taken  account  of, 
however.  If  to  the  line  a,  c,  you  add  the  proportion  of  its  length  equal 
to  the  ratio  of  the  crank,  to  the  connecting  rod,  in  this  case  1/6,  and 
subtract  the  same  amount  from  the  line  a,  d,  and  then  draw  the  regular 
curve  e,  f,  g,  through  the  points  thus  found  and  through  the  center  of 
the  line  a,  b.  Then  vertical  distances  to  the  line  e,  f,  g,  will  represent, 
to  the  scale  chosen,  the  force  of  inertia  at  every  point  of  the  stroke. 

The  application  of  the  forces  of  inertia  to  cause  the  proper  action  of 
an  engine  was  first  worked  by  a  practical  man  having  little  or  no  book 
learning.  The  theory  and  method  of  measuring  and  estimating  such 
forces  has  since  been  so  simplified  that  a  child  may  understand  it,  if 
he  will  try. 


Q.   33.      (1897-8.)      What  is  the  principle  of  the  parallelogram  of 
forces? 

Ans.  33.  If  two  forces  are  represented  in  intensity  and  direction  by 
two  lines,  the  resultant  of  said  forces  will  be  represented  in  intensity 
and  direction  by  the  diagonal  of  a  parallelogram  of  which  the  lines  rep- 
resenting the  first  two  forces  are  the  sides.  If  the  resultant  of  two 
forces  is  represented  by  a  line,  the  forces  themselves  will  be  represented 
by  the  adjacent  sides  of  a  parallelogram  formed  with  the  first  mentioned 
line  as  a  diagonal. 

The  formula  for  the  stored  work  in  a  moving  body,  which  is  called 
the  fundamental  formula  of  mechanics,  and  the  formula  for  obtaining 
the  centrifugal  force  of  a  body  should  be  remembered.    The  similarity 
between  these  two  formulae  will  help  to  fix  them  in  the  memory. 
The  formula  for  stored  force  is 
WV2 

=  foot  Ibs. 

64.4 

The  formula  for  centrifugal  force  is: 
WV2 

=  Ibs.  of  centrifugal  force. 

32. 2R 


Q.  40.  (1897-8.)  What  is  meant  by  the  flashing  point  of  an  oil? 
How  may  the  flashing  point  of  a  sample  be  determined? 

Ans.  40.  The  flashing  point  is  that  temperature  of  the  oil  at  which 
it  gives  off  vapor  at  a  sufficiently  high  rate  to  form  an  inflammable 
mixture  with  the  air  above  it. 

The  point  may  be  determined  by  gradually  raising  the  temperature 
of  the  oil  and  occasionally  passing  a  flame  above  it.  The  temperature 
at  which  the  vapor  takes  fire  is  the  flashing  point. 


Q.  44.  (1897-8.)  How  is  the  safe  torsional  strength  of  a  round  shaft 
of  machine  steel  calculated? 

Ans.  44.  The  force  multiplied  by  the  lever  arm  at  which  it  acts 
(the  moment)  is  equal  to  12,000  times  the  cube  of  the  diameter  of  the 

205 


shaft.    The  units  are  in  inches  and  pounds.  Let  P  =  the  force  in  pounds. 
A  =  the  length  of  the  lever  arm  in  inches, 
d  =  diameter  of  the  shaft  in  inches  and  fractions  thereof. 
This  would  be  expressed  arithmetically  as: 

P  =  A  12000  X  d  X  d  X  d. 
Algebraically: 

PA  =  12000d3. 

For  ordinary  working,  about  one-eighth  of  this  value  would  be  taken: 
PA  =  1500d3. 

One  will  find  a  variety  of  answers  to  Q.  44,  which  may  weaken  his 
faith  in  this  kind  of  calculation.  Experience  in  the  use  of  the  formula 
will  certainly  restore  his  confidence.  The  varieties  of  values  given  are 
probably  owing  to  a  difference  of  Opinion  as  to  what  is  safe.  Sometimes 
one  only  wants  a  part  to  last  a  comparatively  short  time,  and  some- 
times it  does  not  matter  much  whether  a  part  breaks  or  not.  In  these 
cases  one  might  wish  to  allow  a  greater  strain  than  in  other  cases. 
We  have  used  machine  parts  subject  to  a  strain  75%  greater  than  the 
largest  given.  These  have  stretched,  bent,  or  broken,  more  or  less, 
sooner  or  later,  however. 

The  modern  understanding  of  the  nature  of  metals  is  that  they  will 
always  give  way  sooner  or  later  under  repeated  strains.  When  a  metal 
part  will  break  is  merely  a  question  of  time.  The  further  the  strains 
are  below  the  elastic  limit  the  longer  the  part  will  last.  It  is  for  this 
reason  that  a  considerable  factor  of  safety  should  be  taken  in  most 
cases.  We  believe  the  best  statement  of  the  theory,  and  digest  of  expe- 
rience, relating  to  the  so-called  "Fatigue  of  Metals"  is  to  be  found 
in  Prof.  Weyrauch's  book  translated  by  Prof.  A.  J.  DuBois.  We  think 
it  is  published  by  Wiley. 


Q.  45.     (1897-8.)     If  the  rule  is  taken  that  a  shaft  must  not  twist 
more  than  one   degree   in   10   ft.,  what  is  the  allowable   moment,   or 
torque,  for  a  machine  steel  shaft  1%"  in  diameter? 
.     Ans.  45.     For  soft  steel  and  wrought  iron,  the  following  rule  may 
be  taken: 

The  allowable  moment  is  equal  to  160  times  the  fourth  power  of  the 
diameter  of  the  shaft.  Inches  and  pounds  are  used. 

This  is  expressed  arithmetically  as: 

160  X  d  X  d  X  d  Xd. 
Algebraically  as: 

160d4  =  M. 

With  a  1%"  shaft,  the  allowable  moment  under  the  given  conditions 
is: 

160  X  1.5  X  1.5  X  1.5  X1.5  =  160  X  (1.5)«=160  X  (5.0625)  =810 
statical  inch-pounds. 

The  formula  given  will  be  found  to  correspond  very  closely  with 
practice.  If  tempered  spring-steel  is  used  the  constant  185  should  be 
used  instead  of  160. 

If  one  wishes  to  know  how  many  degrees  a  given  force  upon  a  lever 
arm  of  a  given  length  will  spring  a  shaft  10  ft.  long  of  a  given  diameter 
he  can  calculate  it  by  the  following  formula: 

M 


160d4. 

in  which  D  is  the  angle  in  degrees,  M  is  the  moment  (i.  e.  the  force 
multiplied  by  the  -lever  arm),  160  a  constant  for  wrought  iron  and  soft 
steel,  and  d  the  diameter  of  the  shaft  in  inches.  For  other  lengths  of 
shaft  the  angle  will  be  proportional  to  the  length. 


206 


Q.  46.  (1897-8.)  If  it  is  assumed  that  the  extension  of  a  rod  is 
always  proportional  to  the  load,  what  load  would  extend  a  machine 
steel  rod  of  one  square  inch  cross-section  its  own  length?  What  is  this 
number  called? 

Ans.  46.  About  30,000,000  Ibs.  This  number  is  called  the  coefficient 
or  modulus  of  elasticity. 

The  utility  of  the  coefficient  of  'elasticity  referred  to  is  be- 
lieved to  be  that  it  obviates  the  necessity  of  taking  into  account  the 
length  of  the  particular  rod.  This  is  illustrated  in  Ans.  47.  This  coeffi- 
cient is  quite  constant  for  different  specimens  of  steel,  but  varies  some- 
what, especially  between  hard  and  soft  metal.  The  First  German  asso- 
ciation, No.  15,  of  Ohio,  find  it  to  be  29,200,000;  your  committee  has 
found  it  greater  than  30,000,000  in  tempered  steel. 


Q.  47.  (1897-8.)  If  a  machine  steel  rod  is  immovably  attached  at 
its  ends  when  it  is  at  a  temperature  of  100°  what  will  be  the  strain  upon 
it  when  it  has  cooled  down  to  50°? 

Ans.  47.  If  1°  fall  in  temperature  contracts  a  rod  .0000065  of  its 
length,  a  fall  of  50°  would  contract  it  .000325  of  its  length.  If  the  rod 
was  prevented  from  contracting  it  would  exert  a  force  equal  to  that 
which  would  extend  the  rod  that  distance.  That  is  30,000,000  X  .000325 
= 9750  Ibs.  per  square  inch. 


Q.  55.  (1897-8.)  A  round  wrought  iron  rod  3  ft.  long  and  %"  diam. 
is  rigidly  fastened  at  one  end  and  extends  horizontally,  what  weight 
will  it  support  at  the  outer  end? 

(b)  What  weight  can  be  put  on  its  outer  end  for  safe  working 
under  ordinary  conditions  of  practice? 

Ans.  55.  (a)  The  greatest  weight  the  rod  could  be  expected  to 
hold  may  be  found  by  the  following:  — 

Rule: — Multiply  the  cube  of  the  diameter  of  the  rod  (in  inches  and 
fractions  thereof)  by  the  constant  number  4900  and  divide  by  the  length 
of  the  rod  in  inches. 

The  cube  of  .75  is  .422;  .422  X  4900  =  2068,  and  this  divided  by  36, 
the  length  of  the  rod,  is  57.8,  which  is  the  largest  weight  in  pounds 
that  the  rod  could  be  expected  to  hold  at  its  end. 

(b)  The  largest  weight  the  rod  could  be  expected  to  hold  without 
being  permanently  bent  may  be  found  by  the  following:  — 

Rule: — Multiply  the  cube  of  the  diameter  of  the  rod  by  1960  and 
divide  by  the  length  of  the  rod. 

This  would  give  about  23  Ibs.  as  the  weight  that  would  bend  the 
rod  to  its  limit  of  elasticity.  About  half  this,  or  say  12  Ibs.,  would  be 
taken  in  practice  for  ordinary  constructions. 


Q.  56.  (1897-8.)  (a)  A  wrought  iron  rod  of  rectangular  cross-sec- 
tion on  1.5  "vertically  and  %"  broad,  is  rigidly  fixed  at  one  end  so  as  to 
extend  horizontally,  what  weight  will  it  support  4.5  ft.  from  the  fixed 
end?  (b)  What  weight  will  it  safely  carry  under  ordinary  working 
conditions? 

Ans.  56.  (a)  The  rule  for  obtaining  the  greatest  weight  the  rod 
could  be  expected  to  hold  may  be  taken  as:— Multiply  the  breadth  by 
the  square  of  the  height  and  this  product  by  the  constant  number 
14000,  and  then  divide  this  last  product  by  the  length  of  the  rod  in 
inches. 

Thus  .75  X  1.5  X  1.5  X  8311  -4-  54  =  260  Ibs.,  as  the  greatest  weight 
the  rod  could  be  expected  to  sustain. 

(b)  The  weight  that  will  "strain  the  rod  to  its  limit  of  elasticity 
may  be  found  by  the  following:  — 

Rule:— Multiply  the  breadth  of  the  rod  by  the  square  of  its  height 
and  this  product  by  3350  and  divide  by  the  length  of  the  rod  in  inches. 
In  this  case  the  result  would  be  104  Ibs.  as  the  weight  that  would  stretch 

207 


the  rod  to  the  limit  of  elasticity.    About  half  this  last  result  is  taken 
in  practice.     This  would  give  52  Ibs. 

In  rods  subjected  to  a  bending  force  (Ans.  55-56)  the  upper  and 
lower  fibers  may  be  overstrained  while  the  rest  of  the  rod  is  within 
the  allowable  stress.  In  special  experiments  for  the  purpose  of  this 
article  the  round  rods  bent  and  stayed  bent  when  subjected  to  a  force 
a  little  greater  than  given  by  the  first  rule.  If  the  force  had  been 
continued  they  would  have  broken.  Probably  the  rods  would  have 
gradually  bent  in  use  with  a  strain  much  greater  than  given  by  the 
second  rule. 

Q.  59.  (1897-8.)  What  is  the  relative  strength  of  wrought  and  cast 
iron — 

(a)  When  subjected  to  a  crushing  force? 

(b)  When  subjected  to  a  tensile  force? 

Ans.  59.  (a)  Wrought  iron  is  from  1/2  to  1/3  as  strong  as  cast  iron 
when  subjected  to  a  compressive  strain. 

(b)  Wrought  iron  is  from  2  to  3  times  as  strong  as  cast  iron  when 
subjected  to  a  strain  in  tension. 


Q.  60.  (1897-8.)  There  is  a  wrought  iron  bolt  %"  diam.  having 
ten  U.  S.  standard  threads  to  the  inch  cut  in  it.  What  is  its  breaking 
strength? 

Ans.  60.    About  12  000  Ibs. 


Question  1.  (1898-9.)  Everything  in  the  universe  occupying  space 
is  considered  under  the  general  head  of  Matter.  Explain  in  this  con- 
nection the  difference  or  distinction  made  between  Atoms,  Molecules 
and  Bodies;  also  define  the  relation  of  these  terms  to  what  are  known 
as  elementary  and  compound  substances. 

Ans.  1.  A  substance  which  by  no  process  can  be  decomposed  or  sep- 
arated into  anything  different  from  itself  is  classed  in  chemistry  as  an 
element;  in  a  scientific  sense  an  atom  is  the  final  or  indivisible  particle 
of  such  a  substance. 

Compound  substances  are  always  composed  of  two  or  more  different 
"elementary"  atoms;  thus  one  atom  of  oxygen,  when  combined  with  two 
atoms  of  hydrogen,  forms  a  single  molecule  of  water,  hence  the  molecule 
is  the  smallest  particle  of  a  compound  substance  that  can  exist. 

A  "mass"  or  tangible  particle  of  any  substance  may  be  referred  to  as 
a  "body." 

It  should  be  understood  that  atoms  and  molecules  are  so  small  as 
to  be  far  beyond  definite  analysis.  Some  idea  of  the  minuteness  of 
matter  can  be  conceived  by  noting  the  large  area  of  water  over  which 
a  single  drop  of  oil  will  distribute  itself.  In  so  thin  a  film  as  this  the 
molecules  of  oil  are  far  from  being  isolated,  and,  incredible  as  the  ex- 
planation may  seem,  scientists  agree  that  all  "matter"  is  constituted 
or  "built  up"  from  particles  of  infinite  size. 

In  detail,  the  foregoing  subject  embraces  the  atomic  theory,  which 
is  universally  accepted  as  a  satisfactory  explanation  of  all  chemical 
phenomena.  Unless  mentally  well  equipped  with  a  good  conception 
of  this  theory  no  engineer  can  hope  to  understand  the  most  important 
facts  relating  to  furnace  combustion  or  other  chemical  and  physical 
changes  over  which  he  assumes  guardianship. 


Q.  2.  (1898-9.)  Pure  iron  is  known  as  one  of  the  elementary  met- 
als— why  cannot  steel  be  so  considered?- 

Ans.  2.  Iron  of  commerce  is  not  a  strictly  pure  metal,  but  when 
free  from  impurities  it  is  one  of  the  so-called  "elements."  Properly 
combined  with  certain  definite  proportions  of  carbon  tne  internal  struc- 
ture of  iron  assumes  the  characteristics  of  steel,  and  therefore,  being 

208 


the  result  of  a  chemical  combination,  steel  cannot  be  considered  an 

elementary  substance. 


Q.  3.  (1898-9.)  It  is  said  chat  elementary  substances  may  lose  their 
identity,  but  can  never  be  destroyed:— explain  why  this  must  be  true 
and  define  the  difference  between  physical  and  chemical  changes  in 
substances;  also  give  a  few  familiar  examples,  confining  the  subject 
to  the  steam  boiler. 

Ans.  3.  Physical  and  chemical  changes  are  treated  separately  in 
science.  Natural  rtluloscphy  or  "physics"  concerns  phenomena  whereby 
substances  ac^ed  f  non  uo  not  lose  identity  of  composition.  Hardening, 
annealing,  breaking  or  magnetizing  a  piece  of  steel  are  simply  phy- 
sical changes — no  atoms  are  added,  none  are  lost.  The  properties  of 
a  substance  may  vary  as  the  result  of  a  physical  change;  for  instance, 
by  a  continue^  abstraction  of  heat  water  is  converted  from  the  fluid 
to  the  solid  etate.  In  this,  and  also  when  water  is  converted  into 
gaseous  st^am,  the  chemical  "make  up"  (barring  all  foreign  matter,  of 
course)  remains  the  same. 

The  melting  of  one  pound  of  i<?e  will  yield  a  pound  of  water,  as 
will  also  the  condensation  of  a  pound  of  steam. 

The  physical  change  in  water,  from  the  fluid  to  a  gaseous  state,  as 
carried  out  in  the  steam  boilers,  is  due  to  a  chemical  process;  i.  e., 
furnace  combustion.  Carbon,  the  predominating  element  in  coal,  under 
the  influence  of  an  elevated  temperature,  is  released,  atom  by  atom, 
and  when  combustion  is  perfect  each  atom  "pairs  off"  with  two  atoms  of 
oxygen,  the  latter  being  derived  from  the  air  which  is  fed  to.  the  fur- 
nace. Nitrogen  is  set  free,  and  the  oxygen,  together  with  the  carbon, 
now  forms  a  product  called  carbonic  acid  gas.  "Burning,"  therefore, 
is  a  chemical  action  and  a  source  of  heat;  the  fuel — minus  ash  and 
clinkers — disappears,  and  apparently  "matter"  has  been  destroyed. 
True,  the  carbon  and  oxygen  HI  their  new  molecular  relation,  together 
with  the  nitrogen  from  which  the  latter  gas  has  parted  company,  have 
flitted  into  space,  but  by  no  means  is  either  lost;  they  have  gone,  but 
only  to  seek  new  combinations,  and  in  time  they  find  the  affinities 
which  Nature  has  decreed. 

To  plant-  life>  under  the  influence  of  sunlight,  devolves  the  office  of 
again  separating  the  carbon  from  the  oxygen,  and  to  this  source  can 
be  traced  our  present  stores  of  fuel.  Fuel  can  be  burned  as  we  will,  but 
each  atom  thereof  represents  a  part  of  a  universe  in  which  things  are 
muchly  shifted  about,  but  nothing  is  ever  completely  annihilated. 


Q.  4.  (1898-9.)  What  is  vacuum?  Why  is  it  impossible  to  maintain 
such  over  or  in  connection  with  water? 

Ans.  4.  Space  .devoid  of  matter.  From  an  engineer's  standpoint, 
a  vacuum  refers  to  some  chamber  partially  relieved  of  atmospheric 
pressure. 

A  perfect  vacuum,  though  realized,  could  not  be  maintained  over  or 
in  connection  with  water,  because  liquids  give  off  vapor,  which  rapidly 
refills  the  empty  space;  at  a  temperature  of  60°  Fahr.  the  absolute 
pressure  due  to  the  vapor  of  water  is  a  quarter  of  a  pound  per  square 
inch.  A  mercury  gage  would  give  an  indication  of  29.4  inches,  which 
represents  the  best  permanent  "vacuum"  obtainable  at  that  tempera- 
ture. 

At  100°  Fahr.  the  same  gage  would  read  28  inches,  and  so  up  the 
scale  for  increased  temperature  until  at  212°,  the  boiling  point,  the 
vapor  balances  the  normal  pressure  of  the  atmosphere. 


Q.  5.   (1898-9.)     Extension  or  volume  expresses  dimension  and  no 
body  can  be  conceived  that  does  not  possess  length,  breadth  and  thick- 

209 


ness;  neither  can  two  bodies  occupy  the  same  space  at  the  same  time, 
hence— matter  is  impenetrable. 

Extension  and  Impenetrability  are  regarded  as  the  essential  prop- 
erties of  matter;  in  addition  to  this  certain  general  properties  are  ac- 
credited. Name  nine  other  general  properties. 

Ans.  5.  No  particle  or  body  of  matter  can  be  conceived  that  does 
not  possess  length,  breadth  and  thickness;  neither  can  two  bodies  oc- 
cupy the  same  space  at  the  same  time. 

The  essential  properties  of  all  matter,  i.  e.,  extension  and  impenetra- 
bility, are  defined  by  the  foregoing  statement.  Nine  other  general 
properties  accredited  to  matter  are  as  follows: 

Weight:     Due  to  the  force  of  gravitation. 

Mobility:  By  virtue  of  which  the  position  of  bodies  may  be  changed 
by  the  application  of  suitable  force. 

Inertia:  Which  defines  the  persistency  of  bodies  to  retain  their 
state,-  be  it  motion  or  rest. 

Divisibility:  All  matter  can  be  divided  into  distinctly  different 
parts.  There  are  practical  limitations  to  dividing  a  substance;  theo- 
retically the  final  limit  is  assumed  to  be  the  atom  or  molecule,  as 
already  explained.  , 

Porosity:  The  ultimate  particles  of  a  substance,  though  apparently 
dense,  are  supposed  to  be  suspended  in  space,  and  therefore  not  in 
actual  contact.  All  matter  is  more  or  less  porous. 

Compressibility:  Which  means  the  volume  of  any  body  may  be 
diminished;  this  is  a  sequence  to  "porosity." 

Expansibility:  The  converse  of  compressibility,  i.  e.,  substances 
will  vary  in  volume  or  size  at  different  temperatures. 

Indestructibility:  Matter  may  be  caused  to  vary  in  form,  but,  as 
already  explained,  its  elementary  existence  can  never  be  destroyed. 

Elasticity:  Within  certain  limits,  and  varying  in  different  sub- 
stances, bodies  tend  to  recover  their  original  shape  when  relieved  from 
the  strain  due  to  an  applied  force. 


Q.  6.  (1898-9.)  Why  are  hardness  or  elasticity  considered  as  spe- 
cific properties  of  matter? 

Ans.  6.  Hardness:  When  used  to  define  the  property  of  a  substance, 
like  stone;  or  elasticity,  when  used  to  characterize  the  peculiar  ten- 
dencies of  rubber,  are  regarded  as  expressive  of  specific  properties;  for 
not  all  bodies  are  hard  or  elastic  in  the  sense  here  implied. 


Q.  7.  (1898-9.)  Affinity,  Cohesion  and  Gravity  are  known  as  natural 
forces.  Explain  by  familiar  examples  some  of  the  effects  attributed  to 
each. 

Ans.  7.  Affinity,  cohesion  and  gravity  are  the  three  great  forces  of 
Nature. 

Affinity  is  the  strongest  of  the  forces,  but  acts  only  through  in- 
finitesimal distances — that  is  to  say,  it  is  the  power  that  binds  to- 
gether the  atoms  which  constitute  the  molecule  of  a  compound  sub- 
stance. Water  exists  because  the  atoms  of  hydrogen  and  oxygen  are 
held  together  by  "affinity."  Affinity  is  considered  an  atomic  force. 

Cohesion  is  weaker  than  affinity — acts  through  greater  but  still  in- 
sensible distances  only.  It  binds  into  a  mass  molecules  of  similar  na- 
ture and  is  the  power  by  which  a  homogeneous  body  retains  its  form, 
tenacity,  etc. 

Cohesion  is  considered  a  molecular  force. 

Gravity  is  the  weakest  of  these  three  natural  forces,  but  acts  through 
all  known  distances.  It  tends  to  bind  "bodies"  together,  as  is  mani- 
fest by  the  attraction  which  is  known  to  be  mutual  between  all  bodies 
in  the  universe. 

Gravity  is  known  as  a  molecular  force. 

219 


Q.  8.  (1898-9.)  Give  another  and  more  familiar  name  for  the  force 
known  as  "adhesion";  give  the  cause  for  its  action  and  examples  of 
what  would  happen  if  it  ceased  to  exist. 

Ans.  8.  Adhesion  is  classed  as  a  force  causing  molecules  of  dif- 
ferent kinds  of  matter  to  "cling"  together.  Chalk  clings  to  the  black- 
board; dust  of  any  kind  clings  to  every  substance;  bricks  adhere  to 
mortar,  etc.  Friction  is  a  form  of  adhesion,  and  it  is  a  mistake  to 
assume  that  the  resistances  due  to  this  cause  are  without  compensat- 
ing advantages.  The  stability  of  the  major  portion  of  everything 
erected  by  man  depends  on  friction,  and  "pell  mell"  destruction  would 
follow  in  the  wake  of  a  frictionless  era. 


Q.  9.  (1898-9.)  What  are  the  three  states  of  matter?  Name  the 
force  which  causes  a  change  of  state. 

Ans.  9.  Matter  exists  in  three  different  states,  which  are  distin- 
guished as  the  "solid,"  "liquid"  and  the  "gaseous"  state.  The  state  of 
matter  varies  with  the  force  of  "cohesion,"  which  is  strongest  in  solids, 
weakest  in  liquids,  and  totally  lacking  in  gases.  The  intensity  of 
cohesion  is  influenced  by  temperature. 


Q.  10.  (1898-9.)  "Force" — define  this  term  in  the  broad  sense  in 
which  it  is  used  in  science. 

Ans.  10.  "Force,"  in  a  broad  sense,  is  "that"  which  causes  any 
sort  of  change  in  matter,  or  such  as  may  vary  the  form  of  substance 
or  alter  its  condition  or  position  in  space. 


Q.  11.  (1898-9.)  Mechanics  is  a  branch  of  science  that  treats  of  the 
effects  of  force  upon  matter.  Nothing  is  known  of  force  except  through 
matter;  the  two  may  be  regarded  as  inseparable.  Motion  is  one  of  the 
effects  of  force;  name  another. 

Ans.  11.  Mechanically  considered,  "force"  is  the  cause  of  motion, 
or  is  "that"  which  tends  to  produce  motion,  either  retarding  or  totally 
preventing  the  movement  of  bodies  to  which  it  is  imparted. 

The  effect  of  force,  not  manifest  as  motion,  is  "strain." 


Q.  12.    (1898-9.)  .  What  causes  "strain"  and  how  is  it  measured? 
Ans.  12.     Strain   is  due  to   resistance  or  the   reaction   of  opposed 
forces.     It  is  measured  in  units  of  weight. 


Q.  13.   (1898-9.)     What  is  motion? 

AnS.  13.  "Motion"  signifies  change  of  position.  A  body  moving  or 
changing  its  place,  with  regard  to  some  point  that  is  fixed,  is  said  to 
have  an  "absolute  motion."  Relative  motion  refers  to  the  movement 
of  a  body  when  taken  with  reference  to  another  moving  point. 


Q.  14.   (1898-9.)     What   is    "velocity"   and    how   is   it   usually     ex- 


Ans.  14.  Velocity,  usually  expressed  in  feet  per  second,  or  some 
convenient  multiples  of  either  of  these  units,  is  indicative  of  the  "rate 
of  motion,"  i.  e.,  time  and  space. 


INTRODUCTION   TO    QUESTIONS   15   to   30    (1898-9). 

The  foregoing  questions  deal  primarily  with  the  well  established 
theories  concerning  the  constitution  of  matter  and  should  have  served 
the  purpose  for  which  these  questions  are  intended,  i.  e.,  to  give  the 
uninformed  a  partial  insight  as  to  how  scientists  view  nature  in  the 

211 


abstract.  A  better  comprehension  of  the  Atomic  Theory,  the  salient 
points  of  which  are  referred  to,  is  another  of  the  objects  sought,  and 
in  no  instance  can  the  aims  of  the  committee  fall  short  of  the  mark, 
where  associations  or  individual  members  have  taken  up  the  work  in 
the  manner  and  spirit  prescribed. 

A  thorough  understanding  of  nature's  laws,  such  as  may  be  ob- 
tained by  an  intelligent  interpretation  of  fundamental  points,  is 
deemed  essential  and  it  is  hoped  that  none  seeking  enlightenment  con- 
sider themselves  so  well  equipped  in  a  "practical  way"  as  to  be  exempt 
in  this  particular. 

In  thus  advocating  the  cause  of  theoretical  knowledge  and  insisting 
that  the  progressive  engineer  or  mechanic  cannot  afford  to  be  blind  to 
the  teachings  of  science,  we  may  say  further,  that  the  most  important 
and  useful  of  these  facts  are  not  at  all  abstruse  and  can  readily  be 
acquired  by  energetic  believers  in  the  forcible  principle  of  removing 
an  obstacle,  in  preference  to  being  continually  hampered  by  it.  The 
second  paper  of  the  Review  is  offered,  therefore,  as  additional  aid  in 
"smoothing  the  roadway."  The  Educational  Committee  feels  confident 
that  its  efforts  in  this  direction  are  not  misdirected. 

In  pointing  out  material  which  should  be  mastered,  to  insure  ad- 
vancement, we  cannot  overlook  the  importance  of  mathematics  as  a 
factor  concerned  in  nearly  every  proposition  of  a  mechanical  nature. 
To  beginners  in  this  most  needful  branch  of  learning  we  would  say 
that,  in  this  as  in  every  other  branch  of  science,  certain  fundamental 
principles  must  be  absorbed  to  facilitate  progress.  Ordinary  arithmetic 
is  the  stepping  stone  to  such  higher  "planes'  as  algebra,  etc.;  for  the 
present,  however,  we  would  advise  a  sojourn  on  the  lower  level,  each 
individual  to  remain  there  until  such  time  that  he  can  "reason"  him- 
self to  a  more  exalted  position  in  mathematics. 

A  good  practical  knowledge  of  arithmetic  means  more  than  an 
ability  to  add  and  subtract,  multiply  and  divide;  more  especially  if 
these  operations  are  carried  out  in  the  mechanical  manner,  which  is 
not  at  all  uncommon.  The  impracticability  of  reaching  this  subject 
by  any  other  method  has  prompted  the  measure  we  provide;  the  article 
which  follows  is  intended  for  all  who  choose  to  inquire  into  their 
mathematical  standing;  the  view  being  taken  from  a  strictly  elemen- 
tary standpoint.  The  matter  is  pertinent  at  this  time  and,  while  di- 
rected at  the  rudiments  of  arithmetic,  its  broader  aim  is  to  inculcate  the 
need  of  constant  reasoning  in  all  that  pertains  to  the  attainment  of  an 
education. 

MATHEMATICS. 

Mathematics  is  the  science  of  quantity  and  embraces  Arithmetic, 
Algebra,  Geometry,  etc. 

Arithmetic  is  more  particularly  the  science  of  numbers  and  to  a 
partial  consideration  of  the  rudiments  of  this  subject  the  following  is 
directed: 

The  first  proposition  engaging  the  attention  of  a  beginner  in  arith- 
metic is  usually  stated  thus:  "A  unit  is  one,"  a  single  thing.  An  en- 
gine, for  instance,  is  built  up  of  distinct  pieces,  but  considered  collec- 
tively, we  refer  to  it  as  a  single  thing.  "One"  as  a  unit,  therefore,  may 
represent  either  a  definite  quantity  or  an  aggregate  of  quantities. 

Quantity  is  anything  which  can  be  increased  or  diminished.  Mag- 
nitude is  quantity,  considered  in  an  individual  form.  Multitude  is 
quantity,  when  made  up  of  individual  or  distinct  parts. 

Number  is  a  term  covering  broadly  all  answers  to  the  question: 
How  many?  When  one  or  more  things  are  "counted,"  the  result  is 
expressed  by  a  number  which  denotes  the  sum  total  of  units  under 
consideration. 

Numbers,  applied  to  any  particular  thing,  as  one  engine,  two  boilers, 
etc.,  are  defined  as  "concrete  numbers";  all  numbers  not  associated  with 
something  tangible,  are  considered  "abstract." 

212 


Notation  is  the  shorthand  method  of  expressing  numbers  by  charac- 
ters called  "figures."  Numeration  is  the  art  of  reading  numbers  so 
written. 

The  figures  commonly  used  are  0-1-2-3-4-5-6-7-8-9;  combinations  of 
these  serve  to  express  any  number  the  mind  can  conceive. 

The  figure  9,  appearing  alone,  stands  for  "nine" — not  because  the 
symbol  is  especially  adapted  to  represent  this  number,  but  because  it  is 
T)art  of  a  "system,"  which  assigns  a  universally  recognized  value. 

When  appearing  alone  9  stands  for  the  greatest  possible  number  of 
units  that  can  be  given  expression,  by  any  single  figure;  counting  from 
one  to  nine  exhausts  the  significant  figures,  and  at  this  point  the 
elasticity  of  the  system  becomes  manifest. 

Further  progress  in  notation  is  made  by  assuming  the  next  higher 
number  as  the  initial  unit  of  a  higher  order,  i.  e.,  one  unit  of  the  new 
order  being  the  equivalent  of  ten  single  things  as  originally  conceived. 
To  express  this  number  we  again  revert  to  the  figure  1,  using  in  con- 
nection therewith  the  auxiliary  digit  0,  writing  10,  to  represent  ten,  and 
considering  same  as  a  single  unit  of  the  second  order  or  place.  The 
need  of  the  character  0,  referred  to  as  an  auxiliary  and  usually  called 
naught  or  cipher  is  self  evident;  it  has  no. assigned  value,  but  its  posi- 
tion at  the  right  of  any  figure  increases  the  numerical  value  thereof 
tenfold. 

At  99  we  have  nine  units  of  the  order  of  tens  and  nine  simple  units, 
the  sum  being  equivalent  to  ninety-nine,  the  largest  number  capable  of 
expression  with  two  figures.  It  is  plain,  however,  that  the  process 
of  building  up  additional  orders  can  be  repeated  and  the  system  ex- 
tended indefinitely. 

Numbers  written  with  more  than  three  figures,  are  for  convenience 
divided  into  groups  of  three  places  each,  which  are  known  as  periods. 
This  grouping  of  the  orders  into  sections  of  three  is  indicated  by  dots, 
placed  to  precede  the  first  figure  of  each  period. 

The  first  period  consists  of  Units,  Tens,  Hundreds,  and  is  called 
Unit  Period;  thus  the  number  583  stands  for  five  hundred  and  eighty- 
three  units. 

The  second  period  is  also  taken  to  consist  of  Units,  Tens  and  Hun- 
dreds, but  each  unit  here  has  a  value  one  thousand  times  greater  than 
assigned  the  figures  in  the  first  period;  hence  583,000  is  read  five  hun- 
dred and  eighty-three  thousand. 

Additional  periods  are  successively  Millions,  Billions,  Trillions  and 
so  on  upward  into  a  maze  of  figures  that  seldom,  if  ever,  need  invade 
the  engineer's  mind. 

The  foregoing  epitome  of  the  Arabic  System  of  Notation  embraces 
points  more  dormant  than  new;  it  is,  however,  the  "essence"  of  arith- 
metic and,  as  such,  the  several  definitions  must  not  suffer  rusty  repose 
in  our  intellects. 

Four  fundamental  operations  in  arithmetical  computations  are  Addi- 
tion, Subtraction,  Multiplication  and  Division,  represented  and  indi- 
cated respectively  by  the  familiar  signs  +,  — ,  X,  -f-.  The  several 
processes  are  based  upon  and  carry  out  the  principles  set  forth  in  the 
ground  work  of  Arabic  notation  and  when  a  correct  idea  of  the  system 
of  "unit  values"  is  once  firmly  rooted  in,  and  thoroughly  comprehended, 
a  mathematical  stride  of  no  mean  importance  has  been  accomplished. 
*  *  *  * 

To  profit  by  the  experience  of  others,  and  to  arrive  at  conclusions 
for  nimself,  also  to  lessen  the  labor  involved  in  many  calculations,  the 
engineer  must  learn  to  read  abbreviated  mathematical  propositions  and 
should  understand  the  "short  cut"  methods  which  facilitates  the  opera- 
tions indicated.  Ihis  does  not  necessarily  imply  a  profound  knowledge 
of  Algebra,  as  the  operations  indicated  in  some  of  the  most  useful 
engineering  formulae  are  readily  solved  by  arithmetical  computation. 
The  fact  that  a  person  can  add,  subtract,  multiply  and  divide  is  not 
enough  for  the  purpose;  there  is  much  more  than  this  between  the 

213 


Covers  of  a  common  school  arithmetic  which  may  be  absorbed  and  used 
with  profit  and  effect. 

Mathematics  is  essentially  a  science  of  reasoning  and  the  better  the 
cultivation  of  the  reasoning  power  the  less  burdensome  becomes  the 
mathematical  labor  attached  to  the  work.  To  this  end,  therefore,  a 
plea  is  made  for  a  recognition  of  the  true  value  and  importance  of  the 
principles  as  set  forth  in  any  ordinary  arithmetical  text  book.  It  may 
seem  tiresome,  or  may  be  even  frowned  upon  as  "folly"  by  some,  to 
again  take  up  studies  of.  the  kind  here  recommended,  Many  of  course 
are  far  past  this  stage,  but  none  have  traveled  a  better  or  easier 
route.  C.  H.  F. 

Note. — Anything  assumed  as  One  may  be  taken  as  a  Unit.  Some 
particular  thing,  or  rule,  established  to  fix  a  permanent  value,  becomes 
a  "standard";  and  as  such  it  serves  as  the  accepted  measure  for  the 
purpose  it  is  designed  for.  The  absolute  need  of  unchangeable  "stand- 
ards," by  which  values,  etc.,  may  be  measured  and  compared,  is  self- 
evident.  Every  civilized  nation  defines  its  units  of  measurement  with 
accuracy  and  upholds  and  enforces  them  by  legal  enactments. 

This  paper  is  largely  given  to  the  consideration  of  the  units  em- 
braced on  the  engineer's  "yard  stick,"  and  those  not  already  conversant 
with  the  subject  should  cultivate  that  intimacy  which  its  importance 
demands.  • 

Q.  15.  (1898-9.)  First  Law — Every  body  continues  in  a  state  of  rest, 
or  of  uniform  motion  in  a  straight  line,  unless  acted  upon  by  some 
external  force. 

Second  Law — Motion,  or  a  change  of  motion,  is  proportional  to  the 
force  impressed  and  is  in  the  direction  of  the  line  in  which  that  force 
acts. 

Third  Law — Action  and  re-action  are  always  equal  and  are  in  oppo- 
site directions. 

Who  was  the  first  to  give  expression  to  the  foregoing? 

Ans.  15.  The  three  "Laws  of  Motion"  are  accredited  to  Sir  Isaac 
Newton. 

Q.  16.  (1898-9.)  Weight  is  due  to  the  force  of  gravitation,  hence 
every  operation  of  "weighing"  is  a  measurement  of  that  mutual  attrac- 
tion, which  is  inherent  to  every  particle  of  matter  in  the  universe. 

Give  the  units  of  weight  commonly  used  in  engineering. 

Ans.  16.  Avoirdupois,  or  Commercial  Weight,  is  the  common  stand- 
ard of  the  United  States;  the  principal  unit  is  the  "pound,"  but  great 
weights  are  more  conveniently  expressed  in  tons,  the  latter  being  a 
multiple  of  the  original  unit.  The  legal  definition  of  the  ton  is  2,240 
pounds;  the  short  ton  of  2,000  pounds  is  customary  in  many  branches  of 
trade. 

The  avoirdupois  ounce  is  1/16  of  a  pound;  in  engineering  calcula- 
tions it  is  preferable  to  express  such  fractions  in  decimals  of  a  pound. 

"The  standard  avoirdupois  pound  is  equivalent  to  the  weight  of 
27.7015  cubic  inches  of  distilled  water,  Weighed  in  air,  at  39.83°  Fahr., 
barometer  at  30  inches." — Haswell. 

The  French,  or  Metric,  system  is  also  a  legal  standard  in  this  coun- 
try. 

Q.  17.   (1898-9.)     A  line  is  length  without  breadth  or  thickness. 

A  straight  line  is  the  shortest  distance  between  two  points. 

How  is  the  measurement  of  length  ordinarily  expressed  and  upon 
what  is  the  unit  based? 

Ans.  17.  The  foot,  and  its  subdivision,  the  inch,  are  the  units  gen- 
erally used  for  expressing  "Long  Measure." 

The  United  States  and  British  standards  are  the  same,  and  the  funda- 
mental unit  is  the  yard,  which  is  said  to  have  been  based  originally 

214 


on  the  length  of  a  pendulum  vibrating  seconds  at  the  level  of  the  sea, 
in  the  latitude  of  London,  in  a  vacuum,  with  Fahrenheit  thermometer 
at  62°.  The  length  of  such  a  pendulum  is  supposed  to  be  divided  into 
39.1393  equal  parts,  called  inches,  and  36  of  these  inches  were  adapted 
as  the  standard  yard. 

Q.  18.  (1898-9.)  As  a  matter  of  convenience  many  of  the  standard 
units  of  measurement  are  either  subdivided  or  used  in  multiple;  thus 
"Time"  is  noted  in  seconds,  minutes,  hours,  etc.  It  would  be  awkward 
to  refer  to  600  seconds  of  time,  but  six  seconds  is  a  more  ready  ex- 
pression than  six-sixtieths  of  a  minute  and  so  on. 

Frequently  certain  units  are  combined  and  thus  form  a  new  meas; 
ure. 

Assuming  the  minute  as  the  unit  of  time,  what  important  other  unit 
is  evolved  when  expressed  in  connection  with  those  representing  weight 
and  distance? 

Ans.  18.  Using  the  minute  as  expressive  of  the  "time";  the  pound 
representing  "weight,"  and  the  foot  indicating  the  space  or  distance 
traversed,  gives  us  the  "foot-pound,"  or  the  unit  of  work.  Time, 
weight  and  space  are  inseparably  linked  in  the  measurement  of  power, 
and  the  elements  embraced  in  this  most  important  unit  should  be  under^ 
stood  in  a  way  inspiring  a  full  confidence  as  to  what  they  really  por- 
tend. Hazy  views  on  this  point  will  impede  progress. 


Q.  19.  (1898-9.)  Explain  the  relation  between  "foot-pounds"  and 
"horse-power." 

Ans.  19.  33,000  foot-pounds  are  the  equivalent  of  one  horse-power; 
both  are  units  for  the  measurement  of  work  or  power — the  relation 
between  them  being  similar  to  that  of  the  pound  and  the  ton;  that  is. 
one  being  a  multiple  of  the  other,  both  expressing  the  same  thing. 


Q.  20.  (1898-9.)  Area  is  a  term  used  to  express  the  superficial  con- 
tents of  any  figure.  Give  the  units  usually  used  in  this  connection 
and  define  a  "plane"  surface. 

Ans.  20.  Surface  or  area  is  expressed  in  square  inches  or  square 
feet,  as  best  befits  the  case.  By  a  "plane  figure"  is  meant  any  flat  sur- 
face enclosed  by  real  or  imaginary  lines  which  form  the  outlines  or 
boundary  thereof.  Such  a  figure  is  assumed  to  have  length  and  breadth, 
without  thickness,  and  its  surface  is  computed  in  squares  correspond- 
ing to  the  units  used  in  measuring  its  lineal  dimensions. 

Q.  21.  (1898-9.)  Length,  breadth  and  thickness  are  essential  to 
volume.  How  would  you  express  the  volume  or  cubical  contents  of 
a  solid? 

Ans.  21.  As  indicated  in  the  question,  three  dimensions  are  essen- 
tial to  volume — that  is,  we  measure  length,  breadth  and  thickness  and;: 
express  "cubical  contents"  in  cubic  inches,  cubic  feet,  and  so  on,  as 
the  case  may  require.  Awkward  and  ridiculous  blunders  are  a  frequent 
result  from  the  careless  confounding  of  the  expressions  noted  in  this 
and  the  preceding  question. 

Q.  22.  (1898-9.)  Give  the  name  of  a  unit  commonly  used  in  measur- 
ing fluids  and  state  upon  what  it  is  based. 

Ans.  22.  The  United  States  standard  gallon,  containing  231  cubic 
inches,  is  a'common  measure  for  liquids. 

Q.  23.  (1898-9.)  Sketch  diagram  representing  blocks,  measuring  in 
inches,  12"  X  12"  X  12";  24"  X  12"  X6";  36"  X  12  3  ;  calculate  cu- 
bical contents  of  each  in  feet,  and  surface  in  square  feet.  Demonstrate 
thoroughly  the  difference  between  solid  and  surface  measurement. 

Ans.  23.     See  cuts. 

215 


-fatf    j  Sfevre  />. 
ANSWER  NO.  23. 


Q.  24.  (1898-9.).  Geometrically  considered  the  circumference  of  a 
circle  is  a  curved  line,  all  points  of  which  are  equally  distant  from  a 
certain  point  within,  called  the  center. 

The  unit  lor  measuring  angles  is  derived  from  the  circumference  of 
a  circle. 

Give  the  recognized  name  of  an  angle  measuring  89  degrees  59  min- 
utes and  60  seconds;  explain  briefly  how  this  measurement  is  applied 
and  the  signs  used  to  abbreviate  the  units  mentioned. 

Arts.  24.  Circular  measurement  assumes  the  periphery  of  the  circle 
divided  into  360  equal  parts,  each  of  the  spaces  being  known  as  a 
"degree."  The  degree  is  divided  into  60  equal  parts,  called  "minutes," 
and  a  further  subdivision  of  the  minute  into  60  parts  gives  seconds. 
The  signs  or  abbreviations  used  are  thus:  89°,  is  read,  eighty-nine  de- 
grees; 59',  indicates  fifty-nine  minutes,  and  60",  refers  to  sixty  seconds. 
An  angle  measuring  89°,  59',  60"  is  the  equivalent  of  90  degrees,  and  is— 
360 

=4 

90 

therefore  the  fourth  part  of  a  circle.     In  more  ordinary  language,  such 
an  angle  is  a  square,  and  geometrically  it  is  defined  as  a  "right  angle." 


Q.  25.  (1898-9.)  Temperature  is  also  expressed  in  "degrees,"  but  it 
is  evident  that  heat  and  cold  do  not  enter  into  a  geometrical  proposition. 

Explain  the  mercurial  thermometer  and  Fahrenheit's  scale. 

Ans.  25.  The  principle  involved  in  the  ordinary  mercurial  ther- 
mometer hardly  requires  special  mention;  interest  centers  rather  to 
the  scale  or  graduation  by  which  the  expansion  and  contraction  of  the 
mercury  is  measured  and  by  which  the  instrument  is  read. 

"Fahrenheit"  is  the  common  U.  S.  standard  in  engineering  practice, 
and  this  scale,  which  takes  the  name  of  its  originator,  marks  the  boiling 
point  of  water  by  212°  and  the  freezing  point  by  32°. 

Zero,  or  0°,  was  fixed  at  the  temperature  acquired  by  a  mixture  of 
ice  and  salt.  Thermometric  scales  are  entirely  arbitrary,  but  once 
established  and  recognized,  this  fact  in  no  way  detracts  from  their 
general  usefulness. 


Q.  26.  (18989.)  Explain  briefly  the  principle  embodied  in  a  barom- 
eter; state  how  this  instrument  is  read  and  what  the  reading  indicates. 

Ans.  26.  The  barometer  is  an  instrument  in  which  the  known 
v.-eight  of  a  column  of  mercury  is  placed  in  opposition  to  that  of  the 

216 


atmosphere,  the  prime  purpose  being  to  ascertain  the  varying  pressure 
of  the  latter.  The  readings  are  taken  in  inches  of  mercury  which 
represents  the  difference  in  levels  between  the  surfaces  of  the  mercury 
contained  in  the  ciste-n  and  the  indication  as  noied  in  the  tube.  It 
should  be  understood  that  the  upper  end  of  the  tube  is  sealed;  also 
that  the  empty  space  in  the  upper  part  thereof  is  a  vacuum,  and  that 
the  pressure  of  the  atmosphere  is  acting  on  the  surface  of  the  mer- 
cury into  which  the  lower  end  of  the  tube  is  immersed.  Knowing  the 
weight  of  a  cubic  inch  of  mercury,  and  also,  being  conversant  with 
the  manner  in  which  pressures  are  transmitted  by  liquids  and  gases 
we  may  readily  recognize  the  principle  which  is  involved  and  also  un- 
derstand how  the  reading  of  the  barometric  inches  can  be  converted  into 
pounds  pressure  per  square  inch. 


Q.  27.  (1898-9.)  What  impression  should  the  reading  of  a  pressure 
-gauge  convey  to  the  engineer? 

Ans.  27.  The  reading  of  the  pressure  gauge  should  convey  to  the 
engineer  a  realization  that  the  force  indicated  in  pounds  pressure  per 
square  inch  is  acting  with  that  intensity  upon  all  the  surfaces  or  areas 
which  are  accessible  to  the  pressure  under  notice  of  the  gauge.  It 
should  not  escape  us,  however,  that  considered  in  the  sense  of  causing 
rupture — the  ends  of  a  string  must  be  pulled  in  opposite  directions  to 
produce  a  strain  equal  to  the  force  applied  at  one  of  the  ends. 


Q.  28.  (1898-9.)  In  a  concise  yet  comprehensive  manner  explain 
the  value  of  the  British  Thermal  Unit  of  heat  and  the  connection  and 
importance  of  this  unit  in  the  measurement  of  heat  exchange. 

Ans.  28.  The  British  Thermal  Unit  of  heat,  usually  abbreviated  to 
B.  T.  U.,  is  a  measure  which  defines  heat  as  a  quantity  irrespective  of 
temperature.  The  unit  is  based  on  and  is  equivalent  to  the  amount  of 
heat  required  to  raise  one  pound  of  water  through  one  degree  of  Fahren- 
heit. 

Similar  quantities  of  different  substances  require  varying  quanti- 
ties of  heat  to  produce  the  same  temperature;  the  B.  T.  U.  therefore 
establishes  a  basis  for  intelligent  comparisons  and  is  used  in  computing 
all  questions  relating  to  the  transmission  or  absorption  of  heat. 

There  is  believed  to  exist  a  definite  relation  between  heat  and 
mechanical  energy;  this  relation  expressed  in  foot  pounds  has  been 
determined  experimentally  and  is  known  as  the  "Mechanical  Equivalent 
of  Heat"  that  is,  778  foot  pounds  of  work  done,  being  regarded  as  the 
equivalent  of  one  d.  T.  U. 


INTRODUCTION  TO  QUESTIONS  31  TO  42    (1898-9). 
ANOTHER  STEP  IN  MATHEMATICS. 

The  value  and  need  of  good,  sound  arithmetical  knowledge,  ranging 
upward  from  the  simplest  propositions,  has  been  accorded  proper  recog- 
nition in  these  columns.  The  fact  that  it  is  quite  possible  to  "get 
along"  without  such  further  embellishment  of  the  science,  as  are  em- 
braced in  the  higher  branches  of  mathematics,  has  also  been  conceded. 

This  admission,  while  pleasing,  perhaps,  to  those  preferring  to 
evade  the  more  advanced  methods  of  calculating,  must  be  qualified, 
however,  by  the  further  assertion,  that  the  "kind"  of  progress  which 
is  possible  under  such  limitations,  may  be  likened  to  the  difference  in 
travel,  between  the  stage  coach  of  old,  and  the  "flying"  express  trains, 
now  so  familiar  in  modern  railroading.  By  either  method,  of  course, 
the  traveler  starts  with  the  expectation  of  reaching  his  destination, 

217 


but  where  the  latter  mode  is  available,  it  is  safe  to  say,  that  usually 
the  business  may  be  done  and  time  enough  remains  to  forget  about  the 
transaction,  before  the  ancient  form  of  conveyance  could  be  "due"  at 
the  further  end  of  tfie  route. 

For  covering  short,  easy  distances,  the  simple  and  more  primitive 
vehicle  continues  to  fit  conditions  not  requiring  the  excessive  elabora- 
tion of  a  railway  system;  so  with  propositions  of  a  mathematical  na- 
ture— easy  problems  are  best  solved  by  simple  methods,  but  when  the 
elements  of  the  case  are  quite  intricate,  the  question  resolves  itself  into 
one  of  two  things,  i.  e.,  a  solution,  acquired  by  a  tedious,  slow  and 
laborious  process,  or  the  same  results,  reached  with  ease,  rapidity  and 
dispatch. 

As  the  railway  emanated  from  the  older  and  slower  methods  of 
transportation,  so  are  the  higher  mathematics  evolved  from  the  funda- 
mental reasoning  on  which  arithmetic  is  based;  both  are  representative 
of  progress,  and  as  the  business  people  look  to  the  former  as  the  means 
of  facilitating  trade,  so  should  progressive  engineers  view  the  latter,  for 
algebraic  expressions  will  continue  to  be  used  by  advanced  writers,  and 
to  expect  anything  but  this  is  fully  as  unreasonable  as  a  demand  for 
a  general  return  to  customs  and  practices  which  are  regarded  as  ob- 
solete. 

It  would  be  a  difficult  matter  to  answer  satisfactorily  "how  far" 
the  stationary  engineer  should  pursue  mathematics,  or  to  say  definitely 
how  little  is. necessary  to  serve  his  purpose;  each  individual,  rather, 
should  estimate  his  needs,  and  referring  more  particularly  to  such 
engineers,  not  favored  with  an  early  training,  the  deficiency  should 
be  made  good  as  fast  as  it  is  found  that  a  lack  of  such  knowledge  acts 
as  an  impediment  to  progress. 

Personal  experience  is  one  of  the  best  ways  to  gain  knowledge— but 
when  dependent  on  this  alone,  progress  is  slow,  because  life  is  too 
short  to  acquire  all  things  by  actual  contact;  advancement  follows  with 
more  certainty  and  greater  rapidity  to  those  who  aim  to  couple  with 
their  own  practice  the  facts  garnered  by  the  multitude  of  other  patient 
workers  in  the  same  field. 

To  distribute  accumulated  experience  is  the  object  of  the  mechanical 
press;  it  is  the  object  of  all  literature  which  pertains  to  the  work  of  the 
engineer  or  the  artisan,  and  is  the  prime  object  of  this,  and  other, 
associations  organized  for  the  dissemination  of  technical  knowledge. 

To  profit  by  reading,  it  is  evident  that  the  language  used  should 
"strike  home,",  as  published  rules  and  deductions  are  usually  expressed 
algebraically,  it  follows  that  the  reader  should  be  conversant  with  the 
mathematical  signs  ordinarily  used  and  should  also  be  familiar  with 
simple  equations,  for  seldom  do  the  most  useful  formula  exceed  the 
scope  of  this  ready  method  of  stating  concisely  the  mathematical  opera- 
tions to  be  performed  on  the  quantities  which  represent  the  elements 
entering  into  the  problem. 

As  expressed  verbally,  or  in  written  language,  every  rule  recites  the 
arithmetical  operations  to  be  performed  with  certain  known  quantities, 
in  order  that  the  value  of  some  other  may  thereby  be  determined. 

Let  us  assume,  now,  that  we  are  charged  with  the  duty  of  investi- 
gating the  strength  of  a  double  riveted  lap  joint  and  let  the  mathe- 
matical reasoning  concerned  therewith  serve  to  make  plain  the  opera- 
tions usually  involved  in  any  formula,  when  set  forth  in  the  form  of  a 
simple  algebraic  equation. 

In  this  particular  case,  the  elements  entering  into  this  problem  are 
as  follows: 

The  tensile  strength  of  the  plate  per  square  inch  of  section,  the 
thickness  of  plate  in  inches  and  the  portion  of  the  plate  remaining 
unimpaired  between  rivet  holes;  the  product  of  these  three  factors 
represents  the  ultimate  strength  of  the  joint,  as  far  as  the  plate  has 
to  do  with  the  question. 

On  the  other  hand,  we  have  two  rivets,  corresponding  to  the  above 
section  of  plate;  these  have  a  certain  shearing  resistance  per  square 

218 


inch  of  suction;   the  combined  area  of  " 

and  multiplied  by  their  shearing  strength,  gives  the  ultimate  holding 

power  of  this  second  element  of  the  joint...       /-•  (j  y  T 

A  concise  writer  would  probably  express  the  foregoing. elements  as 
follows,  representing  each  of  the  several  factors  by. , means  of. -a- single 
letter,  thus:  .  . 

T  =  Tensile  strength  of  pla"te~prer  square  inch.  .       ; 

p  — Thickness  of  plate. 

r  =  Length  of  plate  remaining  between  rivet  holes. 

A  =  Area  of  the  rivets. 

s  =  Shearing  strength  of  rivets  per  square  inch. 

Then,  T  X  p  X  r  =  A  X  s,  is  the  equation,  covering  the  theory  of  the 
riveted  joint;  interpreted,  it  reads  as  indicated  in  the  preceding  para- 
graphs, i.  e.,  the  tensile  strength  of  the  plate  per  square  inch  (T) 
multiplied  by  the  thickness  of  the  plate  (p)  and  again  multiplied^? 
the  length  of  the  plate  remaining  betw'een  rivet  holes  (r)  gives  a  cer- 
tain product,  which  varies  of  course  according  to  the  values  which 
may  be  given  or  assigned  to  each  of  the  several  factors. 

Now  if  this  product  is  found  equal  to  the  value  of  the  area  of  the 
two  rivets  which  serve  to  hold  the  above  section  of  plate  when  their 
combined  area  (A)  is  multiplied  by  (s)  their  shearing  strength  per 
square  inch,  we  may  conclude  that  the  j'oint  is-at  its -best,  for  It  is 
evident  that  each  side,  or  element,  is  affording  exactly  the -same -resist- 
ance to  a  rupturing  force. 

As  written,  the  equation  TXpXr  =  AXs,  is  general,  i.  e.,.J± 
leaves  the  actual  value  of  each  factor  to  be  found  and  substituted  fey 
the  person  making  practical  use  of  the  rule,  hence  if  the  several  dimen- 
sions and  the  strength  of  the  plate  and  ;rivet  are  known,  and  after 
performing  the  operations  indicated,  it  is  found  that  one  is  greater 
or  less  than  the  other,  the. fault  is  not  with  the  equation,  but  with  the 
joint,  for  the  proposition  may  be  such  that  the  products  found; do; not 
maintain  the  required  equality  which  the  theory  of  the -case  demands:;: 

The  investigation  of  a  given  joint,  according  to  the  method  .ovffi- 
lined,  is  simply  "weighing"  the  question  in  the-  algebraic  "balance," 
called  an  equation,  and  in  this  connection,  it  -i»  necessary  'to  firmly  £x 
in  mind  the  true  meaning  of  the  mathematical  sign  of  equality,  thus,-~. 

The  sign  of  equality  means  just  what  its- name1  indicates,  and- -when 
it  appears  in  a  mathematical  proposition,  it  serves  as  an  indication 
that  the  values  of  the  huinbers,  or  -quantities,  between  which  It  is 
placed,  are  equal,  or  the  equivalent  of  one  another^ — just- as  the  pans 
of  the  "old  time"  balance  come  to  a  level  when  both  sides  are  equally 
weighted. 

Equations  are  solved  by  performing  certain  operation  otr  its  terms, 
the  object  being  to  find  the  value  of  unknown  factors  in  .the  problem, 
by  a  method  of  reasoning  directed  against  others  which  have  a;known 
or  assigned  value.  In  a  general  way,  the  -process  may  be  illustrated 
by  assuming  the  quantities  on  each  side  of  the  sign  =,  to  be  weights, 
laying  on  the  pans  of  an  ordinary  beam  balance.  The  weights  being 
equal,  the  pans  stand  at  a  level  and  so  they  will  remain,  provided 
equivalent  measures  only  are  added,  or  taken  from  either  side  at  the 
same  time.  y  . . ;-",-;'" 

Let  us  now  again  take  up  the  general  equation .  noted  in  connection 
with  the  theory  of  riveted  joints;  where  T  X  p  X  r— A  X  s.  Assume 
the  factors  p  Xr,  to  have  a  balancing  value,  the  exact  amount  of  which 
we  need  care  nothing  about.  Simply  to.  illustrate  the  proposition,  how- 
ever, we  say  that  p  X  r  represents  half  the  weight  contained  in  ,th,e,  left 
hand  pan  of  the  balance.  We  can  remove  this  half  'of  the  weight  from 
the  right  side  and  not  destroy  the  equilibrium  of  the  balance,, provided 
the  weight  in  the  left  hand  pan  is  also  divided  or  diminished  by 
"half."  A  little  reasoning  now  makes  evident'the  fact,  that  p  X  r,  taken 
from  the  right  hand  side  of  the  equation  and  used  as  a  divisor  against 

219 


*bft  terms  appearing  on  the  right,  will  change  the  reading  from 

A  X  s 

TX  PX  r  =  AX  s  to  T  = (1) 

p  Xr. 

Further  investigation  will  prove  that  this  operation  has  in  no  way 
disturbed  the  equality  between  the  two  sides;  that  is,  the  pans  of  the 
balance  remain   level;    we  have  gained  an  important  point,  however, 
by  thus  solving  "T"  and  by  similar  reasoning  we  deduce. 
A  X  s 

(2) 


(3) 
T  Xp 
-also, 

T  X  p  X  r 

(4) 

(5) 

It  will  be  noted  the  original  equation  has  undergone  five  variations, 
that  is,  each  factor  has  been  -arranged  to  stand  alone  in  some  one  of 
these,  hence  if  the  value  of  any  four  of  the  five  are  known,  the  proper 
and  corresponding  value  to  be  assigned  to  the  one  which  is  sought,  can 
be  readily  determined  by  performing  the  operations  indicated,  substi- 
tuting for  the  letters  the  numerical  values  which  they  are  representa- 
tive of;  in  other  words,  With  the  aid  of  a  little  mathematical  reason- 
ing, differing  but  slightly  from  the  kind  ordinarily  used,  the  theory 
of  -a  riveted  joint  have  been  converted  into  five  rules.  These  five 
rules  in  this  form,  occupy  but  little  space  and  answer  effectually  every 
question  that  can  be  raised  in  connection  with  the  subject  to  which 
they  pertain. 

If  it  be  a  question  of  plate  thickness,  against  the  other  values  of  a 
joint,  Equation  (2)  is -used  reading  thus:  thickness  of  plate  (p)  equals, 
•(=)  product  of  rivet  area  and  shearing  strength  of  same,  divided  by 
the  product 'Of  tensile  strength  of  plate  and  length  of  unimpaired  plate, 
between  rivet  holes,  and  so  on. 

It  is  not  the  object  of  this  article  to  instill  more  than  a  mere  ele- 
mentary idea  of  the  higher  mathematics,  neither  has  it  been  the  aim 
to  deal  particularly  with  the  subject  of  riveted  joints,  for  it  should 
be  understood  the  principle  explained  has  a  wide  and  general  applica- 
tion. Neither  of  the  foregoing  subjects  can  be  administered  in  a  single 
heroic  dose;  if,  however,  the  importance  of  this  additional  step  in 
mathematics  is  made  more  apparent  by  the  language  here  recorded,  the 
object  in  view  will  be  quite  fully  accomplished.  C.  H.  F. 


Q.  31.  (1898-9.)  Gravity  is  a  force,  acting  at  all  distances  and  tend- 
ing to  attract  mutually,  all  bodies  in  the  universe. 

Give  the  law  of  gravity  and  state  by  whom  it  was  first  enunciated. 

Ans.  31.  Gravity  is  an  attraction  common  to  all  material  sub- 
stances; the  law  of  universal  gravitation  was  discovered  by  Sir  Isaac 
Newton  and  is  as  follows:  "The  force  of  attraction  between  bodies  of 
•matter  is  in  exact  .proportion  to  their  mass  and  inversely  as  the  square 
of  their  distance  aparti" 

The  first  of  these  propositions  is  quite  plain;  according  to  the 
other,  the  attractive  power  diminishes  in  the  same  proportion  as  the 
"square"  of  the  number  which  expresses  the  increase  of  distance.  Thus 
if  the  distance  is  doubled  the  attraction  is  lessened  fourfold,  because 
4  =  2  X  2  and  is  the  square  of  two,  etc. 


Q.  32.  (1898-9.)  What  common  instrument  serves  to  show  the  direc- 
tion in  which  gravity  acts  and  what  does  this  indicate  to  the  mechanic? 

Ans.  32.  The  plumb  line  and  "bob"  is  a  common  instrument  which 
shows  the  direction  in  which  the  force  of  gravity  acts.  To  the  me- 
chanic it  indicates  a  true  perpendicular;  it  is  evident,  however,  that 
no  two  such  lines  can  be  called  strictly  parallel,  but  for  all  the  ordi- 
nary purposes  of  a  mechanic,  they  may  be  so  considered. 


Q.  33.  (1898-9.)  Is  the  force  of  gravity  constant;  i.  e.,  does  the 
weight  of  a  body  vary  according  to  its  position  with  reference  to  the 
surface  of  the  earth? 

Ans.  33.  Gravity  is  constant  in  the  sense  that  its.  force  is  always 
acting.  "Weight"  is  simply  a  measure  of  the  force  of  gravity  and 
serves  to  indicate  the  intensity  of  attraction  at  or  neaj  the  surface  of 
the  earth  and  at  which  point  is  realized  the  maximum  effect  or  great- 
est weight. 

At  the  earth's  center  weight  is  nullified  and  becomes  manifest  ac- 
cording to  the  following  laws  of  weight: 

Moving  outward  from  the  earth's  center,  weight  increases  as  the 
distance  from  the  center  increases. 

Above  the  surface  of  the  earth  weight  decreases  as  the  square  of 
the  distance  increases.  Therefore,  whenever  the  distance  between 
two  bodies  varies  to  a  sensible  amount,  gravity  must  be  considered  as 
a  variable  force. 

Q.  34.  (1898-9.)  The  "center  of  gravity"  is  a  point  upon  which  a 
body  balances  in  every  position.  What  is  the  difference  between  stable 
and  unstable  equilibrium? 

Ans.  34.  When  a  change  in  position  or  a  slight  displacement  teod& 
to  elevate  the  center  of  gravity  the  body  so  moved  tends  to  return  to 
its  original  position  and  it  is  then  said  to  be  "stable"  or  in  a  state  of 
stable  equilibrium.  When,  however,  the  center  of  gravity-  is  above  the 
point  of  support  and  the  shape  of  the  body  is  such,  that  a  jolt  is  likely 
to  throw  the  line  of  direction  outside  of  its  base,  then  we  have  an  un- 
stable body  or  one  likely  to  fall  because  of  the  constant  tendency  of  the 
center  of  gravity  to  seek  a  position  nearest  to  the  point  of  support. 


Q.  35.  (1898-9.)  Explain  the  difference  between  a  constant  or  uni- 
form rate  of  motion  and  motion  uniformly  "accelerated." 

Ans.  35.  "Uniform  motion"  means,  equal  spaces  described  in  equal 
time;  that  is,  the  "rate"  being  always  the  same.  If  the  body  describes 
a  greater  space  in  each  successive  movement  the  motion  is  "accele- 
rated"; on  the  other  hand,  if  the  spaces  be  less,  motion  is  said  to  be 
"retarded." 

FALLING  BODIES, 

First  law  of  falling  bodies. 

The  space  described  by  a  falling  body,  in  any  given  second,  is  equal 
to  the  product  of  twice  the  number  of  seconds,  minus  one,  times  the 
space  described  the  first  second. 

Thus,  a  body  will  fall  during  the  sixth  second,  viz:  2  X  6=-l£.$i)& 
12  minus  1  =  11. 

Space  described  the  first  second  is  16.08, 

Then  16.08  X  11  =  176.88  feet. 

Second  law  of  falling  bodies. 

The  velocity  acquired  by  a  falling  body  at  the  end  of  any  given 
second  is  equal  to  the  product  of  the  number  of  seconds  into  twice  the 
space  described  the  first  second.  Thus,  the  velocity  attained  at  the 
end  of  the  sixth  second  is  32.16  X  6  =  192.96  feet. 


Third  law  of  falling  bodies.  . 

The  total  space  described  by  a  body  at  the  end  of  any  given  second 
is  equal  to  the  product  of  the  square  of  the  number  of  seconds  into  the 
space  described  the  first  second. 

Thus,  the  total  fall  during  six  seconds  is  16.08  X  36  =  578.88  feet. 


Q.  36.  (1898-9.)  By  whom  were  the  foregoing  laws  determined  and 
under  what  conditions  are  they  true? 

,Ans.  36.  The  "Laws  of  Falling  Bodies"  as  noted  above  were  dis- 
covered by  Galileo. 

The  laws  assume  that  only  the  force  of  'gravity  is  acting  on  the 
falling  bodies;  retardation  due  to  resistance  of  the  air  or  other  influ- 
ences are  not  considered,  as  such  are  always  variable  and  cannot  be 
allowed  for 'in  a  single  b.r'oad  rule. 

Falling  in  a  vacuum  the  lightest  substance  partakes  of  the  same 
rate  of  motion  and  drops  just  as  promptly  as  the  heaviest  metals. 

An  unretarded  falling  body  is  an  example  of  accelerated  motion 
as  may  be  noted  from  the  various  examples  given  in  Q.  36  and  37. 


Q.  37.  (1898-9.)  Knowing  the  laws  of  the  falling  bodies  as  given, 
how  would  you  reason  therefrom  the  final  velocity  acquired  by  a  body 
falling  through  a.  space  of  231  feet? 

Ans.  37.  In  the  absence  of  a  specific  rule,  but  knowing  the  laws  of 
falling  bodies,  an  answer  to  the  proposition  embodied  in  Question  No. 
37  is  obtained  by  reasoning  as  follows: 

Wanted— "the  .final  velocity  acquired  by  a  body  falling  through  a 
space  of  231  feet." 

According  to  the  second  law  of  falling  bodies  the  velocity  acquired 
at  the  end  of  any  given  second  of  time  is  32.16  times  the  number  of 
seconds  and  V  — 32.16  X  t,  is  an  equation  which  expresses  that  fact. 

The  proposition  is  to  find  the  value  of  V,  but  in  order  to  do  this 
it  is  evident -that  a  numerical  substitute  must  be  found  for  "t,"  which 
in  this  case  stands  for  "seconds"  during  which  the  body  is  falling: 

.  From  the  third  law  we  deduce:  h  =  t2  X  16.08,  which  means  "h," 
the  total  height  fallen,  equals  the  product  of  the  square  of  the  time  and 
the  space  traversed  the  first  second. 

Transposing  the  equation: 

f  =  -^-      t  =      /~ 
16.08  -y  ns.08 

Hence  the  value  of  "t"  is  equal  to  the  square  root  of  the  distance 
"h"  divided  by  16.08.  The  expression 


16.08 
can  therefore  be  used  as  a  substitute  for  "t"  in  the  first  equation,  thus: 


For  the  case  in  hand  the  true  value  of  "h"  is  231  and  inserting  this, 
we  have 


* 


16.08 
making  the  value  of  "h"  =  l,  we  establish  a  general  rule,  viz: 

V  =  32'16X-'   16.08 


which  is  equivalent  to 

V  =  8.02  X  l/h" 

Using  this  for  the  present  proposition  gives: 
8.02xl/^T=  121.88  feet. 
So  also  does  the  more  general  rule: 

V  =  l/2gh~=  121.88  feet. 

all  of  which  verifies  the  correctness  of  the  reasoning  and  indicates  that 
principles,  well  fathomed,  are  quite  independent  of  fixed  rules. 


Q.  38.  (1898-9.)  Which  of  the  values  given  in  the  preceding  laws 
do  you  recognize  as  the  "increment  of  velocity"  due  to  gravity?  What 
small  letter  of  the  alphabet  is  generally  used  to  represent  this  value, 
when  such  enters  into  a  proposition  which  is  expressed  algebraically? 

Ans.  38.  32.16  is  a  quantity  well  known  as  the  "increment"  of  ve- 
locity. This  value  is  slightly  variable  for  different  parts  of  the  globe, 
and  it  is  customary  therefore  to  represent  it  by  the  small  letter  "g" 
in  all  general  formulae  in  which  it  occurs;  as  a  factor. 


Q.  39.  (1898-9.)  The  specific  gravity  of  a  substance  is  expressed  by 
figures  which  represent  the  relative  or  proportionate  weight  of  that 
substance  as  compared  with  an  equal  bulk  of  some  other  which  is  fixed 
upon  as  a  standard. 

Name  the  standard  substances  usually  assumed  and  explain  how  to 
determine  the  weight  of  a  cubic  foot  of  metal  when  its  specific  gravity 
is  given  at  7  7/10.  What  metal  is  probably  referred  to? 

Ans.  39.  Water  at  62°  Fahr.,  weighing  62.355  pounds  per  cubic  foot, 
is  the  standard  usually  referred  to,  in  connection  with  specific  gravity. 

For  gases,  the  density  of  atmospheric  air  is  made  the  basis  of  com- 
parison. 

The  weight  of  a  cubic  foot  of  any  substance  may  be  found  by  multi- 
plying its  specific  gravity,  by  the  weight  of  a  like  volume  of  water; 
hence  if  7.7  represents  the  S.  G.  then  62.355  X  7.7  =  480.  The  product 
corresponds  with  the  weight  of  a  cubic  foot  of  wrought  iron. 


Q.  40.  (1898-9.)  What  constitutes  a  "buoyant"  substance  and  what 
law  governs  the  degree  of  submersion  of  a  body  placed  in  water? 

Ans.  40.  Any  solid,  which  weighs  less,  volume  for  volume,  than  the 
fluid  in  which  it  is  immersed  is  said  to  be  a  "buoyant  substance." 

The  law  of  submersion  is  that  an  immersed  body  loses  an  amount  of 
weight,  equivalent  to  the  weight  of  the  fluid  it  displaces. 


Q.  41.  (1898-9.)  State  what  proportion  of  a  timber,  having  a  cross 
section  of  12  by  12  inches,  will  remain  above  water,  when  its  specific 
gravity  is  0.5.  How  much  additional  weight  must  probably  be  added 
for  every  foot  of  its  length,  to  cause  it  to  sink  until  submerged? 

Ans.  41.  The  specific  gravity  of  timber  being  .5,  its  weight  per 
cubic  foot  is  62.5  X  .5  =•  31.25  pounds  or  just  half  the  weight  of  like 
volume  of  water.  In  accordance  with  the  law  quoted  in  the  preceding 
case  and  noting  the  proportionate  weights  of  the  water  and  the  timber, 
it  is  evident  that  one-half  of  the  latter  will  remain  above  water  and 
if  its  section  is  12"  X  12"  an  additional  load  of  31.25  pounds  on  every 
foot  of  its  length  will  be  required  to  balance  the  "buoyant  effort"  due 
to  keeping  the  timber  down  on  the  water's  surface. 


Q.  42.     (1898-9.)     Explain  the  instrument  known  as  the  Hydrometer. 

Ans.  42.  The  hydrometer  is  an  instrument  devised  for  determining 
the  density  of  liquids.  The  graduations  follow  an  arbitrary  scale  of 
degrees  in  Baume's  and  other  instruments.  The  hydrometer  really  indi- 
cates the  specific  gravity  of  fluids  and  its  scale  can  be  arranged  to  read 
that  way  directly,  or  the  arbitrary  scale  readings  in  degrees,  may  be 
converted  into  specific  gravity  by  reference  to  the  proper  tables.  The 
buoyant  power  of  liquids  varies  with  their  weight  or  density  and  this  is 
the  principle  made  use  of  in  the  hydrometer.  The  instrument  is  marked 
at  0  or  zero  when  immersed  in  water  and  the  graduations  running  both 
ways  from  this  point  indicate  a  greater  or  lesser  density  when  the 
instrument  is  immersed  in  the  liquid  which  is  to  be  tested. 


PREFACE    TO   QUESTIONS   OF   1899-1900. 

FALLING   BODIES. 

To  the  progressive  engineer,  a  thorough  knowledge  and  understand- 
ing of  this  subject  is  certainly  advisable.  All  computations  in  relation 
to  bodies  in  motion,  in  their  different  characteristics,  are  affected  by 
these  rules,  and  without  a  knowledge  of  these  principles  one  is  in  the 
dark  why  certain  formulae  and  rules  are  given. 

I  trust  that  a  few  questions  on  the  subject  will  in  no  wise  appear  as 
a  bugbear  to  our  members  who  are  earnestly  seeking  for  a  higher 
standard  of  improvement,  not  only  in  their  manual  training  but  in  their 
social  life,  and  I  trust  that  those  who  are  willing  to  work  will  be 
amply  repaid  for  Uie  time  spent. 

Refer  back  to  last  year's  elementary  course  and  refresh  your  mind 
on  the  treatment  of  the  laws  of  motion,  etc.,  and  then  grapple  with  the 
few  questions  of  this  year.  Some  standard  work,  for  reference,  should 
also  be  in  the  hands  of  all  engineers. 

"Acceleration."  When  an  unrestricted  force,  acting  upon  a  body,  sets 
it  in  motion  (i.  e.,  gives  it  velocity),  in  the  direction  of  the  force,  this 
velocity  increases  as  the  force  continues  to  act,  each  equal  interval  of 
time  (if  the  force  remains  constant)  bringing  its  own  equal  increase 
of  velocity.  Thus,  if  a  stone  be  let  fall,  the  force  of  gravity  gives  to  it, 
in  the  first  inconceivably  short  interval  of  time,  a  small  velocity  down- 
ward. In  the  next  equal  interval  of  time  it  adds  a  second  equal  velocity, 
so  that  at  the  end  of  the  second  interval  the  velocity  of  the  stone  is 
twice  as  great  as  at  the  end  of  the  first  one,  and  so  on. 

We  may  divide  the  time  into  as  small  intervals  as  we  please,  and 
each  such  interval,  the  constant  force  of  gravity  gives  to  the  stone  an 
equal  increase  of  velocity. 

The  rate  of  acceleration  is  the  acceleration  which  takes  place  in  a 
given  interval  of  time,  usually  one  second. 

The  unit  rate  of  acceleration  is  that  which  adds  unit  of  velocity  in 
a  unit  of  time,  or  when  English  measures  are  used,  one  foot  per  second 
per  second. 

For  a  given  rate  of  acceleration  the  total  accelerations  are,  of  course, 
proportional  to  the  times  during  which  the  velocity  increases  at  that 
rate. 

Laws  of  acceleration: 

First — When  the  forces  are  equal  the  rates  of  acceleration  are 
inversely  as  the  masses. 

Second — When  the  masses  are  equal  the  rates  of  acceleration  are 
directly  as  the  forces. 

We  thus  arrive  at  the  principle  that,  in  any  case,  the  rate  of  accelera- 
tion is  directly  proportional  to  the  force  and  inversely  proportional  to 
the  masses.  Hence,  if  we  make  two  forces  proportional  to  two  masses, 
the  rates  of  acceleration  will  be  equal;  or  for  a  given  rate  of  accelera.- 
tion  the  forces  must  be  directly  as  the  masses. 


Time  and  space  in  our  paper  would  not  permit  me  to  extend  this 
study  in  detail,  yet  I  will  call  your  attention  to  some  headings  which 
should  be  carefully  studied  and  thoroughly  understood.  For  my  own 
part,  I  regard  the  laws  of  falling  bodies  one  of  the  most  interesting 
studies  that  I  have  ever  taken  up.  I  will  call  our  attention  to: 

The  constant  force  of  gravity;  the  acceleration  of  gravity;  the  rela- 
tion between  force  and  mass;  the  unit  of  mass  impulse.  Up  to  this 
point  we  should  know  definitely  what  to  Consider  as  the  unit  of  velocity, 
force,  mass  and  time.  From  this  knowledge  we  will  find  that: 

force  X  time 
Vel.  = 


velocity  X  mass 

Force  = 

time 

force  X  time 
Mass  =  — 

velocity 

mass  X  velocity 
Time  = 


•  force 

Also  force  X  time  —  mass  X  velocity. 

This  is  simply  a  problem  of  four  factors,  having  three  of  them  given 
to  find  the  fourth,  which  by  transposition  we  see  are  easily  obtained. 

Forces  in  opposite  directions;  inertia  and  the  densities  of  masses 
should  be  carefully  investigated.  Force  in  relation  to  work  and  so  on 
until  you  arrive  at  the  direct  laws  of  falling  bodies,  which  at  some 
future  time  may  be  fully  explained  to  you. 


Q.  61.  (1899-1900.)  Having  the  temperature  of  sensible  heat  of 
steam  given,  give  the  rule  for  finding  the  total  heat  of  the  steam.  Il- 
lustrate with  an  example  expressed  in  formula. 

Ans.  61.  Using  Kent  for  authority,  the  total  heat  of  saturated 
steam  is  found  by  first  subtracting  from  the  temperature  of  a  given 
pressure,  32°,  and  multiplying  the  remainder  by  .305  (the  specific 
heat  of  saturated  steam),  and  to  the  product  add  1091.7. 

Example: 

When  H=   total  heat,  and  t  =  the  temperature. 

Then  H  =  . 305  (t — 32°).  +  1091.7. 

The  temperature  or  sensible  heat  of  steam  at  70  Ibs.  g.p.  is  316  1° 

The  total  heat  equals  H.  Then,  H  =  .305  (316.1° —  32°)  +  1091  7 
=  1178.3  B.  T.  U. 


Q.  62.  (1899-1900.)  What  is  the  total  heat  of  steam  at  100  Ibs 
gauge  pressure?  Also  the  latent  heat? 

Ans.  62.  The  sensible  heat  of  steam  100  Ibs.  g.p.  is  337.8°  F.,  and 
the  total  heat,  using  the  above  formula  (Kent's)  where 

H  =  .305    (t— 32°)  +1091.7. 

H  =  .305   (337.8—32)  +1091.7. 

H  =  .305   (305.8)  +  1091.7. 

H  =  93.27  +  1091.7. 

H  =  1185.0,  or  the  total  heat  of  steam  at  100  Ibs.  g.p. 

For  the  latent  heat  of  this  given  pressure  (100  Ibs.  g.p.)  using 
"Clark's  formula"  for  finding  the  latent  heat  of  steam  at  any  given 
temperature.  Where  L  =  latent  heat;  t  =  temp,  of  the  steam 

L  =  1092.6— .708    (t— 32°) 

L  =  1092.6— .708   (337.8°— 32°) 

L  =  1092. 6— (.708  X  305.8) 

L  =  1092.6—216.5. 

L  =  876.1  B.T.U.,  or  the  latent  heat  of  steam  at  100  Ibs.  g.p. 

225 


Q.  63.  (1899-1900.)  If  the  steam  in  the  boiler  is  270°  and  the  feed 
water  is  110°,  how  many  units  of  heat  will  be  necessary  to  add  to  the 
water  to  turn  one  pound  of  it  into  steam? 

Ans.  63.  Using  "Kent's  formula"  the  number  of  heat  units  to  be 
added  will  be . 

.305  (t— 32°)  +  (1091.7—110)  =  .305  (270°— 32°)  +  981.7  —  (.305 
X  238)  +981.7  =  72.59  +  981.7  =  1054.29  heat  units  to  be  added. 


Q.  65.  (1899-1900.)  Which  is  the  better  conductor  of  heat,  dry  or 
moist  steam?  Why? 

Ans.  65.  Moist  steam  is  the  better  conductor  of  heat.  Dry  steam 
is  a  poor  conductor  of  heat,  as  compared  with  liquid  water,  or  moist 
steam,  for  after  moist  steam  has  received  enough  heat  to  make  it  dry, 
or  nearly  so,  it  receives  additional  heat  very  slowly. 


Q.  75.  (1899-1900.)  Does  the  change  from  water  to  steam,  by  the 
application  of  heat,  affect  the  relation  of  the  particles  of  the  fluid? 
What  has  this  change  to  do  in  relation  to  power? 

Ans.  75.  As  water,  the  particles  are  strongly  cohesive;  as  steam, 
the  particles  are  repellant.  It  is  this  repellant  force  existing  among 
the  infinitely  small  atoms  of  steam  which  appears  to  give  the  energy 
to  the  mass  of  steam  and  render  it  serviceable. 

The  fluid,  as  water,  is  inexpansive,  but  the  change  to  steam,  by  the 
application  of  heat,  gives  it  energy  or  ability  to  do  work,  by  the  reason 
of  its  great  expansive  or  elastic  tendency. 


Q.  81.  (1899-1900.)  What  letter  of  the  alphabet  denotes  accelera- 
tion? 

What  is  the  rate  of  acceleration  per  second,  in  feet,  of  a  body  falling 
freely  in  vacuo,  from  a  state  of  rest? 

What  is  the  total  acceleration  in  feet  at  the  end  of  one,  two,  three, 
four,  five,  six  and  seven  seconds? 

What  will  be  the  distance  fallen  from  a  state  of  rest  at  the  end  of 
each  second  (as  above  stated)  ? 

What  will  be  the  distance  fallen  (as  above  stated)  in  feet  between 
each  consecutive  second? 

Ans.  81.  Acceleration  is  denoted  by  the  letter  "g"  and  equals  32.2 
feet. 

A  body,  falling  freely  in  vacuo  from  a  state  of  rest,  acquires,  by 
the  end  of  the  first  second,  a  velocity  of  about  32.2  feet  per  second;  and 
in  each  succeeding  second  an  addition  of  velocity,  or  acceleration,  of 
about  32.2  feet  per  second. 

In  other  words  the  velocity  receives  in  each  second  an  acceleration 
of  about  32.2  feet  per  second,  or  is  accelerated  at  the  rate  of  32.2  feet 
per  second  per  second. 

This  rate  is  generally  called  simply  the  acceleration  of  gravity. 
'It  increases   from    about   32.1    feet   per   second    per   second   at   the 
equator  to  about  32.5  feet  at  the  poles. 

These  values  at  the  sea  level,  but  at  a  height  of  five  miles  above 
that  level  it.  diminishes  by  only  one  part  in  400.  For  all  practical 
purposes  it  may  be  taken  at  32.2  feet. 

The  total  acceleration  in  feet  of  a  body  falling  in  vacuo,  from  a 
state  of  rest,  may  be  found  by  the  following  rule: 

Rule: — g  multiplied  by  the  time  the  body  was  falling  equals  the 
total  acceleration  acquired  at  the  end  of  that  second. 

The  acceleration  acquired  end  of — 

1st  second  equals     32.2  feet. 

2d   second  equals     64.4  feet. 

226 


3d    second  equals    96.6  feet. 

4th  second  equals  128.8  feet. 

5th  second  equals  161.0  feet. 

6th  second  equals  193.2  feet. 

7th  second  equals  225.4  feet. 

The  total  distance  fallen  from  a  state  of  rest,  at  the  end  of  each 
second,  is  found  by  the  following  rule: 

Rule: — The  square  of  the  time  of  the  falling  of  the  body  multiplied 
by  V2g. 

A  body  will  fall  at  the  end  of  the — 

1st  second     16.1  feet. 

2d   second    64.4  feet. 

3d    second  144.9  feet. 

4th  second  257.6  feet. 

5th  second  402.5  feet. 

6th  second  579.6  feet. 

7th  second  788.9  feet. 

The  distance  fallen  in  feet  between  each  consecutive  second  is  found 
by  the  following  rule: 

Rule: — g  multiplied  by  the  number  of  seconds  the  body  was 
falling  minus  %  second  equals  the  distance  a  body  will  fall  from  the 
end  of  one  second  to  the  end  of  the  next  succeeding  second. 

A  body  will  fall,  from  rest  in  the — 

1st  second     16.1  feet.' 

2d    second  (or  from  the  end  of  the  first  to  end  of  second) 
48.3  feet. 

3d    second    80.5  feet. 

4th  second  112.7  feet. 

5th  second  144.9  feet. 

6th  second  177.1  feet. 

7th  second  209.3  feet. 


Q.  82.  (1899-1900.)  In  answer  to  this  question  give  the  rule,  or 
state  in  formula,  what  will  be  the  acceleration  acquired  in  a  given 
time  (from  rest) ;  in  a  given  fall  (from  rest) ;  in  a  given  fall  in  a 
given  time  (from  rest)? 

What  will  be  the  time  required  for  a  given  acceleration;  for  a  given 
fall  (from  rest) ;  for  a  given  fall  (from -rest  or  otherwise)  ? 

What  will  be  the  fall  for  a  given  time  (from  reat) ;  required  for  a 
given  acceleration  (starting  from  rest);  during  any  given  second 
(counting  from  rest)  ? 

Ans.  82.    The  acceleration  acquired  in  a  given  time  =  g  X  time. 

In  a  given  fall  (from  rest)  =  V  2  g  X  fall. 

In  a  given  fall  from  rest  in  a  given  time  =  twice  the  fall  -4-  time. 

The  time  required  for  a  given  acceleration  =  accel.  -4-  g. 

For  a  given  fall  (from  rest)  =  V  fall  -4-  y2  g,  or  fall  -=-  %  final  ve- 
locity. 

For  a  given  fall  from  rest  or  otherwise  =  fall  -=-  mean  velocity,  or 
=  fall  -f-  V2  (initial  vel.  +  final  vel.). 

The  fall  in  a  given  time  (from  rest)  =  time  X  Vz  final  vel.,  or  = 
time2  xy2  g. 

For  a  given  acceleration  starting  from  rest  —  acceleration2  -r-  2  g. 

During  any  given  second  counting  from  rest  =  g  (number  of  sec- 
onds— y2  second). 


Q.  83.  (1899-1900.)  Does  the  resistance  of  air  affect  the  strict  ac- 
curacy of  these  rules?  Does  the  specific  gravity  of  different  bodies 
affect  their  falling  properties?  How  do  these  laws  apply  to  bodies 
thrown  upwards  vertically  in  the  air  with  a  given  velocity? 

227 


Ans.  83.  Owing  to  the  resistance  of  the  air  none  of  the  above  rules 
gives  perfectly  accurate  results  in  practice,  especially  at  great  veloci- 
ties. The  greater  the  specific  gravity  of  the  body  the  better  will  be 
the  result. 

The  air  resists  both  rising  and  falling  bodies.  If  a  body  be  thrown 
vertically  upwards  with  a  given  velocity,  it  will  rise  to  the  same 
height  from  which  it  must  have  fallen  in  order  to  acquire  said  velocity; 
and  its  velocity  will  be  retarded  in  each  second  32.2  feet  per  second. 
Its  average  ascending  velocity  will  be  one-half  of  tliat  with  which  it 
started;  as  in  all  other  cases  of  uniformly  retarded  bodies.  In  falling 
it  will  acquire  the  same  velocity  that  it  started  up  with,  and  in  the 
same  time.  

Q.  84.  (1899-1900.)  The  time  a  weight  was  falling  (from  rest) 
was  seven  seconds,  what  was  the  distance  in  feet,  passed  through? 
What  was  the  distance  at  the  end  of  the  fifth  second?  What  was  the 
distance  passed  through  between  the  sixth  and  seventh  second? 

Ans.  84.  The  distance  the  weight  would  pass  through  in  7  seconds 
would  be  d  =  t2  ^>  g  =  72  X  16.1  =  49  X  16.1  =  788.9  feet  ans. 

Second  condition  d  =  t2  %  g  =  52  X  16.1  =  25  X  16.1  =  402.5  feet  ans. 

Third  condition  d  =  g  (7  —  %)  =  32.2  X  6V2  =  209.3  feet  ans. 

In  which  d  =  distance  fallen  in  feet. 

t  =   time  in  falling  in  seconds. 

g  —  acceleration  in  ft.  per  second. 


Q.  85.  (1899-1900.)  What  will  be  the  impact  (striking  force)  in  foot 
pounds  of  a  weight  of  450  pounds  with  a  given  velocity  of  80  feet  per 
second? 

Ans.  85.  The  energy  exerted  equals  the  work  performed.  The  fol- 
lowing rule  describes  how  the  energy  of  a  moving  body  at  a  given  ve- 
locity can  be  determined. 

Rule: — The  weight  of  the  body  in  pounds  multiplied  by  the  ve- 
locity squared,  and  this  product  divided  by  twice  the  rate  of  accelera- 
tion (32.2)  in  feet  per  second  equals  the  work  performed,  or  energy 
exerted. 

Stated  in  formula — 
Wv2        450  Ibs.  X  802          450  X  6400 

F  = =  —  —  =  - —  —  =  44721  ft.  Ibs. 

2  g  64.4  64.4 

Thus  F  =  44721  foot  pounds.    Energy  exerted,  in  which 

F  =  force,  or  energy  exerted. 

W  =  weight  of  the  body  =  450  Ibs. 

v  =  velocity  in  ft.  per  sec.  =  80  ft. 

g=rate  of  accel.  in  ft.  per  sec.  =  32.2  ft. 

By  another  rule  we  find  that  the  measure  of  actual  energy  is  the 
product  of  the  weight  of  the  moving  body  multiplied  by  the  height 
from  which  it  must  fall  to  acquire  its  actual  velocity. 

Let  v  =  velocity  in  ft.  per  sec.  and  h  =  height,  it  must  fall  then 
according  to  the  laws  of  falling  bodies — 
V2          80' 

h  = = =  99.378882  ft,  or  the  height  of  the  fall  to  acquire  a 

2g        64.4 
vel.  of  80  ft.  per  sec.,  then — 

450  Ibs.  X  99.378882  ft.  =  44721  foot  pounds. 

The  same  result  as  obtained  by  the  first  rule,  or  formula. 


Q.  86.     (1899-1900.)     In  raising  a  weight  of  16,000  pounds  by  the 
aid  of  a  screw  jack,  the  screw  %-inch  pitch,  or  four  thread  to  the  inch 

228 


(barring  friction)  how  many  pounds  pull  must  be  applied  to  a  bar  36 
inches  long  from  center  of  screw  to  center  of  pull?  Approximately, 
what  is  the  efficiency  of  a  screw? 

Let  W=weight  in  lbs.=16000.x=power  required,  then  mean  circum- 
ference =  72  inches. 

W:x  ::ir72  :   %. 

Substituting  16000  :  x  :    :   3.1416X72  :   %. 
16000  :  x  :    :  226.2  :   %. 

Then,  as  the  product  of  the  extremes  16000  X  %  =4000,  equal  the 
product  of  the  means  (226.2)  x,  x  =  4000  ^  226.2  =  17.6  Ibs.,  the  re- 
quired power.  Or,  as  the  weight  to  be  raised  is  to  the  power  required 
to  raise  it  in  one  revolution  of  the  screw,  so  is  the  circumference  of 
the  circle  described  by  turning  of  the  lever  to  pitch  of  the  screw. 

In  practice  the  screw  is  used  as  a  combination  of  leverage  with  an 
inclined  plane;  a  spiral  inclined  plane  being  formed  by  the  threads  of 
the  screw.  While  the  power  applied  to  the  lever  which  turns  the 
screw  moves  around  an  entire  circle,  the  body  moves  only  the  distance 
between  the  centers  of  two  threads. 

The  friction  of  the  screw  (which  under  heavy  loads  becomes  very 
great)  has  also  to  be  overcome  by  the  power;  and  this  fact  makes 
these  calculations  of  but  little  use. 


Q.  93.  (1899-1900.)  How  many  pounds  of  steam  at  85  Ibs.  pressure 
per  square  inch  (absolute  pressure)  will  be  required  to  raise  the 
temperature  of  760  Ibs.  of  water  from  58°  F.  to  164°  F.,  the  water  and 
steam  mingling  freely  together? 

Ans.  93.  The  mixture  of  steam  and  water  at  a  temp,  of  164°  F.  con- 
tains 132.404  B.  T.  U.;  at  58°  F.  water  contains  26.007  B.  T.  U.  164°  F. 
—  58°  F.  =  106°  F.,  the  difference  in  temperature  of  the  water  and  the 
steam  and  water  mixed.  132.404  B.  T.  U.  —  26.007  B.  T.  U.  =  106.397 
B.  T.  U.,  the  number  of  B.  T.  U.  in  106°  F.  temp. 

Therefore,  the  number  of  B.  T.  U.  that  each  pound  of  water  will 
absorb  equals  106.397  B.  T.  U.  760  X  106.397  =  80861.72  B.  T.  U.  equals 
the  number  required  to  raise  760  pounds  of  water  from  a  temp,  of  58° 
F.  to  164°  F. 

The  steam  mingling  with  the  water  will  give  up  its  latent  heat  of 
evaporation,  which,  at  a  pressure  of  85  Ibs.  absolute  per  sq.  in.,  is  891.3 
B.  T.  U.  per  pound. 

It  will  also  give  up  a  portion  of  its  sensible  heat  in  falling  from  a 
temperature  due  to  85  Ibs.,  which  is  316°  F.  to  a  temperature  164°  F. 

Steam  at  316°  F.  contains  287.022  B.  T.  U.  (sensible). 

Water  at  164°  F.  contains  132.404  B.  T.  U.  (sensible). 

287.022 — 132.404  =  154.618  B.  T.  U.  that  the  steam  gives  up  in 
addition  to  its  latent  heat  (891.3  B.  T.  U.),  thus  891.3  +  154.618  = 
1045.918  B.  T.  U.  per  pound  of  steam. 

The  number  of  pounds  of  steam  required  is  equal  to  the  number 
of  B.  T.  U.  absorbed  by  the  water,  80861.72  divided  by  the  B.  T.  U. 
given  up  by  one  pound  of  steam,  1045.918. 

80861.72  -f- 1045.918  =  77.31  pounds  of  steam  required  under  condi- 
tions of  question. 

Formula — 

W  (ts  —  ti)  760(164  —  58) 

S  =  —  —  =  —  —  —  77.22  ans. 

L+(t  — 12)         891.3+  (316  — 164) 

In  which — 

S  =  steam   required   in   pounds. 

W  — the  water  to  be  heated,  760  Ibs. 
tj  =  initial  temp,  of  water  =  58°  F. 
t2  =  final  temp,  of  water  =  164°  F. 
t  — temp,  of  the  steam  at  given  press.  316°  F. 

L  =  latent  heat  of  steam  at  given  press.  891.3  B.  T.  U. 
229 


Q.  105.  (1899-1900.)  What  is  understood  by  the  term  "Mechanical 
Efficiency"  of  an  engine? 

By  the  term  "Thermal  Efficiency"  of  an  engine? 

What  is  the  mechanical  efficiency  of  an  engine  developing  500  I.  H. 
P.  and  425  brake  H.  P.? 

What  is  the  thermal  efficiency  of  an  engine  using  steam  at  130  Ibs. 
absolute  pressure,  exhausting  at  6  Ibs.  absolute  pressure  per  sq.  in.? 

Ans.  105.  The  "mechanical  efficiency"  of  an  engine  is  the  ratio  of 
the  actual  horse-power  used  in  performing  the  work,  to  the  indicated 
horse-power,  or  the  mechanical  energy,  developed  in  the  cylinder  real- 
ized in  useful  work  expressed  in  percentage. 

Rule. — Multiply  the  actual  horse-power  by  100  and  divide  this  prod- 
uct by  the  indicated  horse-power  and  the  quotient  will  be  the  mechani- 
cal efficiency,  expressed  in  percentage. 

The  "thermal  efficiency"  of  an  engine  is  the  ratio  of  the  heat  ex- 
pended in  the  cylinder,  is  to  the  total  heat  entering  the  cylinder,  or  the 
total  heat  entering  the  cylinder  that  is  used  in  performing  work. 

425  X  100 

=  85  per  cent  mechanical  efficiency. 

500 

130  Ibs.  absolute  =  347.1°   F. 

6  Ibs.  absolute  =  170.1°  F. 

347.1  +  460  =  807.1,  total  temp,  of  steam  entering  cylinder. 

170.1  +  460  =  630.1,  total  temp,  of  steam  exhausting  from  cylinder. 

807.1  —  630.1  =  177.0°  F.  that  the  temp,  of  the  steam  is  reduced  to 
in  performing  the  work. 

Then— 

177  X  100 

=21.66  per  cent  thermal  efficiency. 

807.1 

Formula — 

T_Ti 

X  100  =  per  cent  T.  E. 

T 

Substituting— 
807.1  —  630.1 

—  x  100  =  21.66  per  cent  of  thermal  efficiency. 
807.1 

T  =  absolute  temp,  of  initial  steam. 
T*  =  absolute  temp,  of  exhaust  steam. 


Q.  21.  (1900-01.)     What  is  the  unit  of  heat  and  how  is  it  expressed? 

Ans.  21.  Unit  of  Heat.— The  British  unit  of  heat,  or  British  ther- 
mal unit  (B.  T.  U.),  is  that  quantity  of  heat  which  is  required  to  raise 
the  temperature  of  1  pound  of  pure  water  1°  at  or  near  39.1°  F.,  the 
temperature  of  maximum  density  of  water. 

The  French  thermal  unit,  or  calorie,  is  that  quantity  of  heat  which 
is  required  to  raise  the  temperature  of  1  kilogramme  of  pure  water  1° 
Cent,  at  about  4°  Cent.,  which  is  equivalent  to  39.1°  F. 

1  French  calorie  =  3.968  B.  T.  U. 

1  B.   T.  U.  =  .252  calorie. 


Q.  22.  (1900-01.)     What  is  the  mechanical  equivalent  of  heat? 
Why  do  we  say  heat  has  a  mechanical  equivalent? 
How  many  foot  pounds  are  represented  by  each  B.  T.  U.,  and  what 
name  is  given  to  this  equivalent? 

230 


Ans.  22.  The  Mechanical  Equivalent  of  Heat. — Is  the  number 
of  foot  pounds  of  mechanical  energy,  equivalent  to  1  British  thermal 
unit?  Heat  has  a  mechanical  equivalent  because  they  both  are  mutu- 
ally convertible. 

Joule's  experiment  (1843-50)  gave  the  figure  772,  which  is  known 
as  Joule's  equivalent.  More  recent  experiments  by  Prof.  Rowland 
gives  higher  figures,  and  the  most  probable  average  is  now  considered 
to  be  778.  That  is,  1  B.  T.  U.  =  778  foot  pounds,  or  778  foot  pounds 
is  considered  as  the  mechanical  equivalent  for  1  B.  T.  U. 


Q.  23.  (1900-01.)  If  one  heat  unit  equals  778  foot  pounds  of  energy, 
what  decimal  part  of  a  heat  unit  does  one  foot  pound  represent? 

How  many  heat  units  per  hour,  per  minute,  per  second,  are  required 
for  one  horse-power? 

One  pound  of  carbon  burned  to  C  Oa  (carbon  dioxide  or  perfect 
combustion)  equals  14,544  heat  units;  what  is  the  efficiency  of  one 
pound  of  carbon  per  horse-power  per  hour? 

Ans.  23.  If  1  heat  unit  equals  778  foot  pounds  of  energy,  1  foot 
pound  equals  1/778  =  .0012852  heat  units. 

1  H.  P.  =  33000  foot  pounds  per  minute. 

1  H.  P.  =  2545  B.  T.  U.  per  hour. 

1  H.  P.  =  42.416  +  B.  T.  U.  per  minute. 

1  H.  P.  =  .70694  B.  T.  U.  per  second. 

1  pound  of  carbon  burned  to  C  02  =  14544  heat  units. 

1  pound  of  carbon  per  H.  P.  per  hour  =  2545  -f-  14544  =  17%  per 
cent  efficiency. 

Q.  25.  (1900-01.)  What  is  thermal  capacity,  and  what  is  the  co- 
efficient of  thermal  capacity? 

Ans.  25.  Specific  Heat. — The  thermal  capacity  of  a  body  is  the 
quantity  of  heat  required  to  raise  its  temperature  1°. 

The  ratio  of  the  heat  required  to  raise  the  temperature  of  a  given 
substance  1°  to  that  required  to  raise  the  temperature  of  water  1° 
from  the  temperature  of  maximum  density;  39.1°  F.  is  commonly  called 
the  specific  heat  of  the  substance,  or  co-efficient  of  thermal  capacity. 


Q.  26.  (1900-01.)  How  is  tne  specific  heat  of  a  substance  obtained 
or  determined  according  to  the  method  by  mixture?  Give  formula 
for  the  same. 

Ans.  26.  Determination  of  Specific  Heat. — Method  by  Mixture: 
The  body  whose  specific  heat  is  to  be  determined  is  raised  to  a  known 
temperature,  and  then  is  immersed  in  a  mass  of  liquid  of  which  the 
weight,  specific  heat  and  temperature  is  known.  When  both  the  body 
and  the  liquid  have  attained  the  same  temperature,  this  is  carefully 
ascertained. 

Now  the  quantity  of  heat  lost  by  the  body  is  the  same  as  the  quan- 
tity of  heat  absorbed  by  the  liquid. 

Let— 

G  =  specific  heat  of  the  hot  body; 

W  =  weight  of  the  hot  body; 

t  =  temperature  of  the  hot  body; 

C*  =  specific  heat  of  the  liquid; 

Wl  =  weight  of  the  liquid; 

t1  =  temperature  of  the  liquid. 

T  =  temperature  the  mixture  assumes. 

Then,  by  the  definition  of  specific  heat,  CXWX  (t  —  T )  =  heat 
units  lost  by  the  hot  body,  and 

231 


C1  X  W1  X  (T  — t1)  =heat  units  gained  by  the  cold  liquid. 
If  there  is  no  heat  lost  by  radiation  or  conduction  these  must  be 
equal,  and 

OW1  (T  — t1) 

C  W  (t  — T)  =CJ  W1  (T  — t1)   or,  C  = 

W  (t  — T) 


Q.  33.  (1900-01.)  How  would  you  place  a  globe  valve  on  a  steam 
pipe  with  the  pressure  above  or  below  the  valve,  and  why? 

Would  you  place  the  valve  so  the  stem  is  in  a  vertical  or  horizontal 
position? 

Ans.  33.  A  globe  valve  should  be  placed  on  a  steam  pipe  with  the 
pressure  below  the  valve.  There  are  opinions  contrary  to  this,  but  it 
is  believed  that  such  opinions  are  entertained  principally  by  cranks 
and  those  who  like  to  be  on  tne  off-side. 

It  is  advocated  that  the  pressure  being  on  top  of  the  valve  that  it 
is  not  so  liable  to  leak.  This  is  erroneous.  If  the  seat  is  cut  a  valve 
will  leak  regardless  of  the  pressure. 

Again,  if  the  valve  becomes  detached  from  the  stem  with  the  pres- 
sure on  top,  operations  have  to  be  suspended  until  the  valve  is  replaced 
or  repaired.  For  convenience  of  packing  the  stem,  the  pressure  should 
be  placed  on  the  under  side  of  the  valve. 

It  is  proper  to  have  the  stems  of  the  valves  horizontal.  This  is 
also  a  matter  of  convenience  for  opening  and  closing  and  also  avoids 
forming  a  water  pocket. 

Valves  so  placed  that  the  stems  hang  down  beneath  the  pipe  is  not 
good  practice,  and  should  be  severely  condemned. 


Q.  35.  (1900-01.)  Describe  the  "steam  loop,"  its  uses  and  upon  what 
does  its  action  depend. 

Furnish  sketch. 

Ans.  35.  The  "Steam  Loop." — It  is  a  system  of  piping  by  which 
water  of  condensation  in  steam  pipes  is  automatically  returned  to  the 
boiler. 

In  its  simplest  form  it  consists  of  three  pipes,  which  are  called  the 
"riser,"  the  "horizontal,"  and  the  "drop-leg." 

When  the  steam-loop  is  used  for  returning  to  the  boiler  the  water 
of  condensation  and  entrainment  from  the  steam  pipe  through  which 
the  steam  flows  to  the  engine  cylinder,  the  riser  is  generally  attached 
to  a  separator;  this  riser  empties  at  a  suitable  height  into  the  hori- 
zontal, and  from  thence  the  water  of  condensation  is  led  into  the  drop- 
leg,  which  is  connected  to  the  boiler,  into  which  the  water  of  conden- 
sation is  fed  as  soon  as  the  hydrostatic  pressure  in  drop-leg  in  connec- 
tion with  the  steam  pressure  in  the  pipes  is  sufficient  to  overcome  the 
boiler  pressure.  The  action  of  the  device  depends  on  the  following 
principles: 

Difference  of  pressure  may  be  balanced  by  a  water  column;  vapors 
or  liquids  tend  to  flow  to  the  point  of  lowest  pressure;  rate  of  flow 
depends  on  difference  of  pressure  and  mass;  decrease  of  static  pressure 
in  a  steam  pipe  is  proportional  to  rate  of  condensation.  In  a  steam 
current  water  will  be  carried  or  swept  along  rapidly  by  friction. 

The  water  of  condensation  runs  into  a  separator.  The  drip  from 
the  separator  is  below  the  boiler,  and,  evidently,  were  a  pipe  run  direct- 
ly to  the  boiler,  we  would  not  expect  the  water  to  return  up  hill. 
Moreover,  the  pressure  in  the  boiler  is  say,  100  pounds,  while  in  the 
separator  it  is  probably  about  95  pounds,  due  to  the  drop  of  pressure  in 
the  steam  pipe,  by  reason  of  which  difference  the  steam  flows  to  the 
engine.  Thus  the  water  must  not  only  flow  up  hill  to  the  boiler,  but 
must  overcome  the  difference  in  pressure. 

232 


The  device  to  return  it  must  perform  work,  and  in  so  doing  heat 
must  be  lost. 

The  loop,  therefore,  may  be  considered  as  a  peculiar  motor  doing 
work,  the  heat  expended  being  radiation  from  the  upper  or  horizontal 
portion. 

From  the  separator  or  drain  leads  the  pipe  called  the  "riser,"  which 
at  a  suitable  height  empties  into  the  "horizontal."  This  leads  to  the 
"drop-leg,"  connecting  to  the  boiler  anywhere  under  the  water  line. 
The  "riser,"  "horizontal"  and  "drop-leg"  form  the  loop,  and  usually 
consist  of  pipes  varying  in  size  from  %-inch  to  2  inches,  and  are  wholly 
free  from  valves,  the  loop  being  simply  an  open  connection  from  sepa- 
rator to  boiler.  (For  convenience  stop  and  check  valves  may  be  used, 
but  they  take  no  part  in  the  loop's  action.) 

Suppose  steam  is  passing,  engine  running  and  separator  collecting 
water.  The  pressure  of  95  Ibs.  at  the  separator  extends  back  through 
the  loop,  but  in  the  drop-leg  meets  a  column  of  water  which  has  risen 
from  the  boiler  when  the  pressure  is  100  Ibs.,  to  a  height  of  about  10 
feet;  that  is,  to  the  hydrostatic  head  equivalent  to  the  5  Ibs.  difference 
in  pressure. 

Thus  the  system  is  placed  in  equilibrium.  Now  the  steam  in  the 
horizontal  condenses  slightly,  lowering  the  pressure  to  94  Ibs.,  and  the 


column  in  the  drop-leg  rises  6  inches  to  balance  it;  but  meanwhile  the 
riser  contains  a  column  of  mixed  vapor,  spray  and  water,  which  also 
tends  to  rise  to  supply  the  horizontal  as  its  steam  condenses,  and  being 
lighter  than  the  liquid  water  of  the  drop-leg,  it  rises  much  faster. 

If  the  contents  of  the  riser  have  a  specific  gravity  of  only  .1  that 
of  the  water  in  the  drop-leg,  the  rise  will  be  ten  times  as  rapid;  and 
when  the  drop-leg  column  rises  1  foot  the  riser  column  will  lift  10  feet. 

By  this  process  the  riser  will  empty  its  contents  into  the  horizontal, 
whence  it  is  a  free  run  to  the  drop-leg  and  thence  to  the  boiler. 

In  brief,  the  above  may  be  summed  into  the  statement  that  a  de- 
crease of  pressure  in  the  horizontal  produces  similar  effects  on  con- 
tents of  riser  and  drop-leg,  but  in  degree  inversely  proportional  to  their 
densities.  When  the  condensation  in  horizontal  is  maintained  at  a 
constant  rate  sufficient  to  give  the  necessary  difference  of  pressure,  the 
drop-leg  column  reaches  a  height  corresponding  to  the  constant  differ- 
ence and  rises  no  higher.  Thus  the  loop  is  in  full  action,  and  will  main- 
tain circulation  so  long  as  steam  is  on  the  system,  and  the  difference  of 
pressure  and  quantities  of  water  are  within  the  range  for  which  the 
loop  is  constructed. 

No  water  should  accumulate  in  the  separator,  as  it  is  the  mission 
ot  the  loop  to  remove  it  before  it  assembles  into  a  liquid  mass. 

233 


It  is  here  that  constant  and  vigorous  action  is  of  great  practical 
utility,  enabling  the  loop  to  act  as  a  preventive  rather  than  a  device 
for  removing  water  after  it  has  accumulated.  The  separator  evidently 
must  be  of  such  form  as  to  give  the  sweep  toward  and  through  the  loop 
better  opportunity  to  pick  up  the  entrained  water  than  is  afforded  by 
the  current  sweeping  toward  the  engine,  pump  or  steam  using  device. 
The  loop  action  is  practically  independent  of  the  distance  the  source 
of  supply  is  above  or  below  the  boiler  and  also  independent  of  the 
length  of  return.  It  is  capable  of  handling  such  quantities  of  water  as 
usually  exist  in  steam  systems.  It  is  practically  limited  by  excessive 
differences  in  pressures,  and  abnormal  quantities  of  water. 


234 


The  National  Association  of  Stationary  Engineers 

ORGANIZED  OCTOBER,  1882.     INCORPORATED  OCTOBER,  1892 


Three  hundred  and  eighty  subordinate  Associations  with  sixteen  thousand  members 
in  forty-eight  States  and  Territories 


PREAMBLE. 

This  Association  shall  at  no  time  be  used  for  the  furtherance  of 
strikes,  or  for  the  purpose  of  interfering  in  any  way  between  its  mem- 
bers and  their  employers  in  regard  to  wages;  recognizing  the  identity 
of  interests  between  employer  and  employe,  and  not  countenancing  any 
project  or  enterprise  that  will  interfere  with  perfect  harmony  between 
them. 

Neither  shall  it  be  used  for  political  or  religious  purposes.  Its 
meetings  shall  be  devoted  to  the  business  of  the  Association,  and  at 
all  times  preference  shall  be  given  to  the  education  of  engineers,  and  to 
securing  of  the  enactment  of  engineers'  license  laws  in  order  to  prevent 
the  destruction  of  life  and  property  in  the  generation  and  transmission 
oi  steam  as  a  motive  power. 

FORMER  PRESIDENTS  AND  YEARS  OF  SERVICE. 

1882.  H.  D.  Cozens Providence,  R.  I. 

1883.  James  G.  Beckerleg   Chicago,  111. 

1884.  James  G.  Beckerleg Chicago,  111. 

1885.  R.  J.  Kilpatrick   St.  Louis,  Mo. 

1886.  F.  A.  Foster Bridgeport,  Conn. 

1887.  G.  M.  Barker Boston,  Mass. 

1888.  R.  O.  Smith New  York  City 

1889.  John  Fehrenbatch Cincinnati,  O. 

1890.  J.  J.  Illingworth   Utica,  N.  Y. 

1891.  William  Powell   Cleveland,  O. 

1892.  C.  W.  Naylor Chicago,  111. 

1893.  James  D.  Lynch Philadelphia,  Pa. 

1894.  M.  D.  Nagle New  York  City 

1895.  Charles  H.  Garlick  Pittsburg,  Pa. 

1896.  J.  W.  Lane  Providence,  R.  I. 

1897.  C.  A.  Collett St.  Louis,  Mo. 

1898.  W.  T.  Wheeler  New  York  City 

1899.  Herbert  E.  Stone Cambridge,  Mass. 

1900.  P.  E.  Leahy  New  York  City 

1901.  E.  G.  Jacques Detroit,  Mich. 


235 


INDEX 

PAGE 

Acceleration             .            .            .            .            .            .            .  226-227 

Adhesion      .             .             .             .             .             .             .             .  211 

Adiabatic  Expansion            .            .            .            .            .            .  .199 

Affinity                 .             .             .             .             .             .             .             .  210 

Air  Compressor  Temperature           .             .             .             .             .  135 

Air  Pump  Capacity         .......  68 

Alternating  Current              .            .            .            .            .            .  .183 

Alternation                      .                         .  "         .            .            .            .  183 

Ampere  .  .  .  .  ...  .  .     145-151-166 

Angular  Advance             .             .             .             .                          .             .  66-128 

Angularity  of  Connecting  Rod         .             .             .             .             .  .        67 

Armature — Energy          .             .             .             .             .                          .  189 

Losses  in  .......      189 

Reaction       .             .             .             .             .             .  189 

Atom               .             .             .             .             .             .             .             .  .      208 

Automatic  Cut-off            .             .             '.                         .            .             .  120 

Pabbitt  Bearing       .             .          '.-  '        .             .             .             .  .        68 

Band  Wheel— Size  and  Weight      .             .             .             .             .  89 

Barometer                   .             .-            .             .  ;          .                          .    .  .      216 

Bearings — Brass  or  Babbitt         .           -.             .             .             .             .  68 

Belts — Improper  Running   .......        70 

Capacity  of           .  '          .             .'""..             .             .             .  98 

Size                  .            .                          .             .             .             .  .98 

Pull  on      .             ...             .             .             .             .  98 

Width— How  to  Join        -  •  >;  -'-'/•'       .            .            .  .        90 

Bo  lers — and  Engine  Efficiency  .  .  .  .       '-'':"         65 

Blow-off      .            .                        .            .'•:'-.            .  4 1-42 

Braces,  Strains  in         .            :.    •        .             .             .             .  5-35 

Bumped  Heads — Pressure            „  -         •.             .             .  .54 

Butt  Straps        .             .            '.          '  .             .             ,             .  53 

Compound,  Universal       .            .            .            .            .  .62 

Corrosion,  External     ......  5 

Dimensions  of                     .            .            .            .            .  .38 

Dome     ........  41 

Factor  of  Safety     ....                          •  •        55 

Grate  and  Heating  Surface  Ratio       ....  6 

Head  of ,  Sketch     .             .             .             .             .             .  10-11 

Heating  Surface  Efficiency     .....  10 

Horse  Power           .             .             .             .             .             .  .9-12 

Horse  Power  Ratios     .        '     .             .             .             .             .  56 

Horizontal,  Support  for     .             .             .             .            .  .44 

Lecture  on         .             .             .             .             .     .     ...  »    •  •  .      .  14 

Metals  Used  in        .             .          •  , :         /.             •             •  •          8 

Mysterious  Gas  in                     .             .             .             .             .  6 

Number  for  a  Plant            .             .             .             .             .  .        40 

Patch  on  Fire  Sheet     .             .             .             .             .             .  10 

Pitch  in  Setting     .......         7 

Pitting                .......  6 

Plate  Formulas       ......  49-52 

"     Tensile  Strength             .            .            .            .            .  14 

"     Thickness      .......        13 

Riveted  Joints — Value             ....  6 

Room  Reports         .             .             .             .             .  .48 

Safety  Valves                .             .             .             .             .             .  7 

Scale  Preventer       .             .             .             .             .             .  -6-7 

Segment  Area  of  .  .  .  .  .  -33-197 

Steam  per  Hour      .....  •        32 

Strains  in           .......  5-9 

236 


Boilers — Surface  Exposed  to  Heat 
"  Supports,  Size  of 

Tests  of 

Test,  Report  of  a 

Tubes  as  Stays 

Tubes,  Iron  or  Steel     . 

Tubular — Construction 
"          Setting 
' '          Specifications     . 

Water  Tube — Advantages 

Where  to  Feed 

Working  Pressure 
Books  Referred  to     . 
B.  T.  U.  .  .  . 

Brushes,  Shifting  of 
Buckeye  Engines  .  . 

Palorimeter— Purpose— Use  of 

"     Carbonic  Acid  Gas— Absorbed 
Carbon  Monoxide,  Percentage  of 
Center  of  Gravity 
Centrifugal  Force 
Chimney — Area         .  . 

"  Capacity 

Draft 


Apparatus 
Specific  Heat 
Temperature 
Weight  of    . 
Waste  in 
' '  Guys  for 

"          Height  of 
"          Size  of  .  . 

"          Steel  or  Iron 
1 '          Temperature 
Circuit — Drop  in  . 

"        Breaker 
"  "        versus  Fuse 

Clearance  in  Engine 
Coal  vs.  Oil  Fuel       . 
Coal — Amount  Required 
"       Combustion  of 
"       Evaporation  from         .    . 
"      and  Combustible 
Combustible  and  Coal     . 
Combustion — 

Flame  from 
' '  Secondary 

of  Coal      .  . 

Steam  Jet      .   ' 
Co-efficient  of  Elasticity 
Cohesion        .  '    . 

Coil  Winding  . 
Compass — Use  of 
Compressed  Air  .  . 

Compression  .  . 

Commuting 
Commutator 
Condenser — Air  Pumps 
Calculation 
Types  of 
"  Water  Required 


PAGE 
44 
36 

15-58 
60 
52 
54 
5 
57 

•  56 
.     6-38-48 

9 
55 

•  36 
202-217-230 

.   186 
69 

.   37-203 

•  30 
3i 

.    221 

204 

•  45 
45 

43-44-46 
29 
30 
29 

29-30 
.  20-29-30 

•  3i 
13 

44-47 
43 
43 

10-23 

147-148 

153 

•  153 
.   70-117 


10 

•       37 

i7 

•  f3 

64 

.    64 

8-27-28 

•  13 
207 

210 

186 

.      163 
24 

•  67 

184 
.  184 

141 
.  140 

140 
.  142 


237 


•     ,A     •  PAGE 

Conductors  in  Series        .......  160 

Connecting  Rod— Shape  of              .  .            .            .            .            .     '  84 

Size  of  .....        84-85 

Strain  on  .....        82 

Corliss  Engine — Speed               .            .  .            .            •            •             75 

"           Valves        ...  .             .             .             .67 

Corrosion              .             .             . ';  .             .             .        -    ,              61 

Crank-Pin  Pressure                ...  .             .             .             .        79 

Crank  Shaft — How  to  Line-up               .  .                          .             .              92 

Pressure         .            ,  .            .            .            .     .       .      101 

Crank  and  Piston  Inertia            .             •  '    -    '.''"    '•.-.'.'"'.              81 

Cross-Head  and  Piston  Inertia         .  .             .             .             .      •       .        81 

Critical  Temperature       .             .  ".  ;          .           \            .            212 

Current — Direction                .            '.  .             .             .             .             .188 

"          High  Tension              .            .  .            ...            146 

"           Potential               ".             .  .             .             .             .             .      146 

Cut-Off—              .                          .             .  .             .             .             .            120 

"          Equal  or  Not         .  .             .             .             .             •        71 

Cycle — Alternating         ...  .             .             .             .            184 

Cylinder — Condensation       .             .  .          •  .,         .  -c                             79-84 

"           Diameter        .             .             .  .             .            ...              71 

"           Low  Pressure — Size  of    .  .                          .             .79 

F)raft  in  Chimney               .             .-  .             .    :                     .           43-44-46 

Dry  Pipe     .  "          .             .             .  .             .             .             .             .  4I 

Dynamo         .                                                _  „                148-187-190-191-192-194-195 

Bi-polar -Multi-polar             .    •        .•    -        .  .             .              153 

"         Brushes       .             .             .  .             .             .                          .       153 

Compound       .            .            .  .       -     .    '        .            .             193 

Capacity     .            .            .  .        '    .    y       .                         .152 

Efficiency         .             .             .-  .           "  .        '     .            .              195 

"         How  Connected     .             .  .             .         '   ."      -     .             .       153 

Shunt  or  Compound               .  .            .       ...             152 

"         Shunt          ;             .             .  ;             .             .             .             .       191 

' '        Series               .            .  •         .•  -    .            .            .       '  '  .             1 92 

vSparking    .             .             .  .             .        :  •'•            .             .       153 

Horse  Power  of            .-            .  .             .             ."         '.              152 

parth — a  Magnet          .             .             ,  .             .             .  .          .              162 

Eccentric  -Turning  Down       ......         66 

Angular  Advance        .  .             .             .             .             .66 

119 

Efficiency  of  Heating  Surface  '     . .           .            .            .            .  iO 

Elastic  Limit            .            .            .  .            .            .            .            .        44 

Elasticity            .             .             .             .  .             .-..,.           210 

E.  M.  F.— To  What  Proportioned  .            .            .            .            .       183 

Electric  Connection,  Rule        .            .  .            .            .            .     ,        154 

"        Conductors,  Danger          .  .            ...            .184 

.            .  .            .  .-         .            .             186 

Current,  To  Produce          .  .         :  .            .          ...            .       187 

"         Developed     .             .  .            .             .             .              159 

Strength               .  .          '  .             .             .             .       178 

Quantity  of               .  .             .             .             .              179 

"         Generated            .  .             .             .            •.             .       180 

"         Direction      ......       182-183 

Circuit — Open  or  Closed   .  .             .             .             .             .160 

"        Non- Conductors          ......  156 

"        Series          .            .            .  .            .                        .            .       156 

Electrical — Induction    .             .             .  .             .             .                    181-182 

Work  Units       .             .  .             .             .             .             .179 


Horse  Power 


144 


Instruments      .......  154 

Energy       .....  195 

Unit 151 

238 


Electrical— Knowledge 

Action   . 

Electricity — Current  of 
"  Static   . 

"  Positive  and  Negative 

"  Definition 

Electro— Static 

"          Dynamic  V 

"          Saturation 
"          Magnet 
Electro-Motive  Series    . 
Electrified  Body — Grounded 
Elevator  Repairs 
Engine  and  Boiler  Efficiency 
Engine — Clearance 

"          Compound,  H.  P.  of 

"          Cylinder  Size 
.   "  "          Advantage 

"          Connecting  Rods 

Corliss        . 

"          Capacity  to  Increase 
"          Cranks 
"          Cross  Compound 
"          Cross  Heads 
"          Cylinder  Wear 
4<  "         Drips 

"         Ratio  .   " 

"          Economy  of      •    *•  . 

* '          Eccentric        .  . 

"          Four- Valve.  Economy     . 
High  Speed  -When  Use 
"          Governor  . 

"  "         Corliss 

* '          Horizontal  or  Vertical     . 

Increased  Load 
' '          Knocking  In        ; 
Multi-Cylinder 
Most  Economical 
"          Mechanical  Efficiency- 
New,  How  to  Start 
"          Receiver— Purpose     . 

' '  Pressure 

"          Rotary 
"          Single-Acting 
"          Size  Required 
"         Set  Up— Cost 
"          Thermal  Efficiency ". 
"          Underloaded 
Engineers,  Famous        .       :"._  . 
Evaporation ,  Efficiency  of 

Equivalent          ; 
From  Coal 
Good 

Temperature  of 
Suspicious 
Extension     .  .          .  ,- 

pactor  of  Safety 

Falling  Bodies— Laws 

Feed  Water— Where  Enter 
Field  Winding — Defects  in 
Firing  Hand 


9.V 


PAGE 

155 

•  155 
159 

i 56- i 59 

•  '55 
155 

•  155 


158 

'  'I 

70-117 
106 
107 


102-108 
103 

.       118 

94-105 

118 

117 


79-94 
H9 

I29 

15-129 

126 

128 

93 

8o. 
130 
106 
117 
230 

92 
104 
105 
116 
129 
87-93 

93 
230 

93 
199 


10 

37 

'         ¥ 

37-6i 

.       209 

55 

221-222-224 
228 


239 


PAGE 

Fly  Wheel — Centrifugal  Force  .  96 

Functions  of     .  .  .  .  .  84- 1 20 

Safe  Speed  .  .  .  .  .  .  80 

Safe  Weights  .  .  .  .  .120 

Construction          .  .  .  .  .  .  120 

vs.  Band  Wheel  .  .  .  .          .  .  .         89 

Pit  .  .  .  .  .  .         .  •  .  91 

Flue  Area      .  .  .  .  .  .  .  .  .         55 

"     Gas  Analysis        .        "     .  .  .  .  .          -;.  17 

Foot  Pounds  and  H.  P.        .  .  .        -  •',  .        .     .  .       215 

Force — Definition          .  .  .  .  .  .  .  211 

Forced  Draft  .  .  .  .  .  .  .  .         43 

Foundations — For  Engine        .  .  .  .  .  .          90-92 

Material       .  .  .  .  .  .  -91 

Template          .  .  .  .          -,.  .          .  91 

Bolts  for       .  .  .  .  ,    -     -  .  .        91 

Capstone  Material         .  .  .  •  *  91 

Wedges  for  ....  .  .  -91 

Frequency  .  .  .  .-  .....  183 

Fuels — Table  of  .  .  .    -        .  .  .28 

Fuses — Diameter  .  .  .  .     -  .  .  149-150 

"     vs.  Circuit  Breakers    .  .  .  .  .;  .       153 

Fusible  Plugs     .  ,  .  .  I  .*.':.  42 

Furnace  Formula      .  .  .  .  .  ,."-'..  .         49 

galvanometer  .  .  ...  .  168 

Gauge — Pressure          .  .  .  .  .         -.  .       217 

Grate— Surface  Ratio  ..  .  .  '         .  .  .  6-40 

"         Distance  of  .  .  .  .         44 

Ratio  to  Chimney        .  .  ..  .  .  .          45-46 

"        Bar  Opening  .  .         .  .  .  .  .  .        55 

Governor  Pulley — Size  of          .  .  .  .  ..\         .         73-128 

Guides — Pressure  on  .  ; .  .  -.  .  .         82-99-101 

Gravity  .  .  .  .  .'  .  .  210-220-221 

~LJ  eat  Utilized  in  Furnace  .  .  .  .  .'  .  29 

"    Used  in  Making  Steam  .  .  .  .  .  29-31 

"    Definition  .  .  .  .  .         '   .      .       .  199 

"    Mechanical  Equivalent  of       .  .  .       ''    -.  .  230 

"    Units  per  H    P.      .  .  .  .  .  .  231 

Heaters,  Open  and  Closed      .  .  .  .  .  .  138 

"        Saving  by  Use    .  .  .  .  .  .  .  139 

Heating  Surface,  Efficiency  .  .  .  .  .  10 

Ratio  .  .  .  .         •   .  .         6-40 

Helix  .  .  .  .  .  .          .  .  .  164-165 

Horse  Power  of  Boiler  .  .  .  .  .  .         9-12 

"          "      and  Cylinder  Diameter  ...  .  .  71 

"          "      to  Raise  Water      .  .  ...  133 

"          "      and  Foot  Pounds         .  .  ..  ...  215 

Hysteresis  .  .  ...  .  .  .  .  189 

Jce  Making  .       .    ..  »  ....  137 

"  Mixture          ........  201 

"  Amount  in  a  Room          ......  133 

Impenetrability  .......  209 

Incandescent  Lamps,  Current  Used  ....  143. 

Incrustation          .  .  .  .  .  .  .  .  61 

Indicator— Card — Sketch      .  .  .  .  '  ,.     .  94 

"      M.  E.P.  .  .  .  .  .  .  112 

"          Card,  Water  from  .  .  .  .  .  .  112 

"          Diagram         .......  109-110 

"          How  Used  ......  108 

"         Theoretical  Curve     .  .  .  .  .  .  in 

240 


PAGE 

Induced  Draft  .                    43 

Inertia  of  Engine  Parts                .  .             .             .             .             .           204 

Injector,  Economy  of            .             .  .-          .             r             .                  134'! 

Size  of  Pipe        .  .             .    •                      .138' 

"        Source  of  Power     .            .  .            '.            .            .                  139 

Injection,  Water  Required          .  ...                         134 

Iron,  Wrought  vs   Cast         .             .  ...      .'.....•               208 

Strength              ..  ...."'.             .            208 

Isothermal  Expansion          .  .            .                  200 

Tack  Shaft           .                          .  .                          .              89 

J  "         "      Diameter            .     '.  .            .            .            .                  119 

Joule         .            .            .  .-           .            .            .            .            178 

Kilo-Watt                   .  144 

patent  Heat  of  Steam                .  .           • .             .             .                  200-201 

Lecture,  Boilers  and  Furnaces  .             .             .             .                    14 

"          Foundation            ...  .             .             .             .86 

Lifting  Water             ...  .             .             .             .           136-137 

Lines  of  Force                   .             ..  .            .             .                162-164-186-190 

Liners  in  Bearings      .             .             .  . .                     •   ..             .                    66 

Link  Motion,  Stephenson           .  .             .             .             .         -   .            125 

Lode  Stone                  .            .            .  .            .            .            .                 161 

Lubricants,  Oils  vs.  Grease         .  '.             .        '    .             .             .            202 

A/Tagnet          .             ....  .             .             .             .                  161 

Electro             .             .  .             .             .             .             .166 

Magnetic  Circuit        .  •           .             .  .             .             .  •          .                  148 

Substance         .             .  .             .                      •    .            .            162 

Field           .  •          .             .  .             .             .                 162-163-188 

Polarity            .             .  .             .             .             .             .            163 

Induction                .             .  .             .             .             .                  163 

Density          ....  .                         .            ,            .            163 

Magnetism      .                          .             .  .                          .             .           160-163 

Permeability             .  .                          ...     149-165 

Residual               .            ».  .             .             .             .                  191 

Mathematics        ...             ;  .             .             .             .             .     212-217 

Matter                           .             .             .  '                     .             .             .                  208 

"        Indestructibility             ......  209 

"       Attributes  of              ......  210 

Three  States  of               .  .             .             .             .             .            211 

Metals,  Thermal  Conductivity     ,    .  .             .             .             .                  199 

Microhm               .             .             .  .             .             .             .             .            171 

Modulus  of  Elasticity            .             .  .             .             .             .                  207 

Molecules          '    .             .             .'  .    '         .             .             .             .            208 

Momentum                  .             .            •'.  .                    •  '  .             .                  198 

Motion                   .             .             .  .             ....             .            2ii 

Laws  of           ...             .  ..           .             .             .                  214 

"       Ratio  of                .             .  .     •      ..             .             .             .            221 

Motor,  Winding  for                ...  ...                  154 

"      Circuits                  .             .  .             .             .             .            154 

Starting  Box           '     ^           •  •             •             «             •                  J54 

"      Cut-out     .             .             .  .            '.          .  .             .            .            154 

'NJ    A.  S.  E.  Educational  Committee  .  -                                             3 

*     '•           Officers,  Past          ...  .             .             .            .             .            235 

Qhm              .....  .          144  151-166-169 

Ohm's  Law  .  .  .  .  .  .  ,.  133-145-167 

Oil  vs.  Coal  Fuel  .  .  .  -.  .  .  .  64 

' '  Viscosity  of  .  ^  .  .  .  ,-  203 

"  Flashing  Point  .  .  .  .  .  .,  205 

Oxygen,  Absorbed           .             .  ...             .            '.             .              30 

241 


PAGE 

parallelogram  of  Forces                    .                          ...  205 

Patch  on  Fire  Sheet              .             .             .             .             .             .  10 

Permeability,  Magnetism                   ...  149 

Pipe,  Expansion  of          .             .             .             .             .                          .  198 

Pitting,  How  Caused             ...                          .  6 

Piston,  Position  and  Velocity                 .             .             .             .             .  81 

"       and  Cross  Head  Inertia                                   .             .  81 

and  Crank  Positions       ......  83 

Speed,  High  .  .115 

Porter  Allen  Engine        .             .             .             .     .        .             .            .  73 

Potential,  Definition                            .             .             .             .           ...  157 

Difference  of                .             .                          ....  146-185 

"         Zero             .             .             .  .          .             .             .             .  157 

Drop  in            .                                                                          .  175 

Pulsating  Current                   .                                       ...  184 

Pumps,  Air,  Size  of         .....                         .             .             .             .  142 

"     Duty  of              .....             ...             .  141 

Capacity              ....  138-139 

Circulating                ...                          .             .  141 

Cylinder  Ratio                             .     •                     .             .            .  138 

Forces  Opposing       ....  138 

Force  Pumps,  Lever      .  135 

Friction         .            .         '   .            .            .            .            .  139 

Head  Against         •         .             .•                     .    .             .             .  135 

H.  P.  Required         .                                       ...  139 

Lifting  Water                  .                                       ...  136-137 

Dimensions                .             .             .       ,      .             .             .  137 

Most  Economical          .            .            .            .  •          .            .  134 

Piston  Speed             .            .            .            .            •.-,.,       .  137 

to  Set  Valves                  .             .             ...          .             .  133 

Slippage         .  .  .139 

Steam,  Cylinder  Diameter        .                                       .  133 

"        Efficiency  of            .            .            ...            .  133 

Successful  Operation                                                         .  138 

Suction  Pipe                           ;                                  ;.-•=...•'        .  138 

Water  Hammer  in         .             .                          .          ;  .        ;     .  134 

.  Radiator  —  Size  required      .             .  •    .     .             .             .  .      203 

Refrigeration     Calculation             .             .             .             .             .  136 

Capacity      ...             .             .             .             .  137 

Rivet— Strength  of  -55 

Diameter  of              .             .             .           ,.            .             .  35~53 

Pitch  of               .             .             .                          .'           .  53 

Riveted  Joint — in  Boilers     .  .....          6 

Pitch  of             .            .            .                        ...  8 

How  Fail     .             .             .             .          ...  -35 

Rod — Horizontal  Strength  of    .                                      .                          .  207 

Rope  Transmission  .         '    .             .   .          .             .             .             .  ,  .        79 

Cafety  Valves      .             .            .             .             .             .             .             .  7- 1 3-42 

Scale— Preventer            .'..•.            .     '       .  .          6 

"         in  Boilers      .             .             .             .                          .             .  12 

"          Loss  Due  to         .              .              ;            n              .              .  -31 

"        When  Deposited      .            .            .                        .            .  200 

Screw  Raising  Weight          .             .             .             .             .             .  .228 

Segment— Area  of           .......  33~'97 

Shaft— Torsional  Strength                .                         .             .        r    .  .      205 
Torque     .             .             .             ...             .             .206 

Shunt  Circuit            .             .  147-160 

Smoke— Burning             .......  8 

Cause  of      ........        63 

Solenoid               ........  165 

Sparking        .  .      153 

242 


Speed  of  Engine             .            .  ,           '„-...           .            .            128 

Spirit  Level— How  Tested   .  ...        92 

Specific  Heat       .                          .  ,                     ,-.'»".     202-231 

"           "     of  Chimney  Gas  .            .            .            .                         .29 

Specific  Gravity               .  .         ^  .             .             .             .             223 

Starting  Box             '.  .  *                      .                          .154 

Stays— Load  on               .             .  .             .             .             .             .               49 

Tubes  as        .  ,             .     -       .             .             .    .         .        52 

Slide  Valve  Cut-off         .  •    . .             .             ...               70 

Steam — Consumption  Indicated  .             .            .             .             .             .        71 

Distribution       .             .  •  .                         ....           1 28 

Drums           .  ...        41 

Expansion  in  Cylinders  .             .             .             .             .      103-104 

Effective  Pressure   .  .             .             .             ,            .             .        88 

Heat  Used  to  Make       .  ,             .                                        31 

Heat  Sensible     •  .                       225-226 

Jacket     .             .             ...  .             .             .         68-91 

Jets  Combustion      .  .            ,            ....        13 

Liquified  by  Pressure    .  .             .             ,     ,        ,             202 

Conductor  of  Heat  .            ,            .            .                  226 

Physical  Condition        .  .             .             .             .             .             226 

To  Raise  Temperature  of  Water    .  .             ...      229 

Loop       .            .            .  .            .            ....            232 

Weight  of    .            .  .            .'           .            . '          .  .          .198 

Required  per  H.  P.       .  .             ...         .             , .       80-88 

Jacket  Results          .  .             .             .          .  .             .             -95 

Used  Expansively          .  .             .             .           . .             .              95 

Pipe  Specification   .  .88 

"     for  Engine             .  .            .            .            .            113 

"     Drips  from       .  .             .                         .                          .114 

Separator  -  Purpose       .  .    •         .             .             .             .             113 

Where  Put  .             .             .            .           ,  .             .114 

Useful  Work  from          .  .            .             .            .            ."'»            33 

Stokers — Mechanical            .  .            .             .            .  -           .             .        44 

Steel— Not  an  Element              ...  .             .             .             208 

Strain             .             .             .  .  •          .             ...      211 

' '      by  Change  in  Temperature  .                                       .           • .             207 

Sublimation               .  •          .  .                          .             ,             .             .      200 

Surface  Blow-off              .             .  .             .             .             .             .              42 

""Temperature — Chimney  Gases  .          .  .             .             .                          .        10 

of  Evaporation  .             .             .             ...              37 

Tensile  Strength— Boiler  Plate  .    /                    .                         ,    ''        .        14 

Ultimate       .  .             .             .    .         ,            v              44 

Thermal— Conductivity  of  Metal  .             .             .•         \      .       .             .      199 

"           Capacity         •',...  •            •             .             •             231 

Theoretical  Curve     .             .        .  .    .         .       "     .             ...            iio-in 

Transformer — Construction        .  ....                         ,  '          146 

Transforming  Direct  Current  .             .             .             .             .                   146 

Terminal  Pressure           .             ...  ".             .             .  .          .  •            71 

Torque            .......  .      188 

Tubes — Expanding  vs.  Beading  .             .             .                          .              52 

Tube  Area      ."''..             .  .                         .                                       -55 

|J nits  of  Area     .            '.    ^       .  .             .             .            ..           .             215 

Circular  Measure  ,             .           ..             .             .             .216 

Length            .     v .     .  -  .                        .            .                        214 

Liquid  Measure    .  .             .             .             .             .             .      215 

Heat    .             .             .  ;.            .             .             .                         230 

Temperature         .  .                          .             .             .             .216 

Time   .             .             .  .             .             .             .             .             215 

Volume     ........      215 

Weight  .  .  .  .  .  -  .  .  .214 

Uptake  Temperature             .  .             .             .             .            .             .8 

243 


PAGE 

Vacuum— What  Is  It?              ,            .  .            .            .  %.      .  .            209 

Valve  Diagram — Zeuner           .  .         '   .  .          ,             .        72-75-120 

"          Bilgram  .            .  .     •    •  ,            .            .         76-77 

"          Sweet            .  .            .            .            .               77~7& 

"     Setting — Corliss        .             -.  .'-''.            .             .              67 

"           Buckeye      ,    .  .             .            ...        69 

"          "           Porter  Allen           .  .             .            ,              73 

Valves  —  Kind  to  Use            .             .  .            ,.            ...            113-232 

Valve  Gear           .             .             .            .  .             .             .             120 

"         "     Functions           .             .  ,  .          .             ...          ,             .       120 

"     Travel         .             .             .             .  .    '                      .66  70-75 

'     Angle  of  Advance       .             .  .             .             ,            .             .121 

"         Admission           .             .             .  .             ,            ...             121 

"        Balanced       .            .            .  .            .        -    .            .            .      121 

"        Compression      .            .            .  .            ,            .            .            121 

"        Cut-off          ,            .  -.            .            .            .           121-125 

Exhaust              .            .        .    .  .           ',            .            .            121 

Lap  .  .           121-124 

Lead        .  ...     121-124 

Gridiron       .                        .  .            .                   •  •  .            .121 

Piston     .  .            .          '  .            124 

"        Poppet          .....  .125 

Relief     .             .           ....  .             .            .121 

"         Rotary           .             ...  .-'-.-.             .             .             .      121 

"         Slide        .             .             .             .  .             .             .             .121 

"        Travel           .            .            .  .            .-.                        .121 

Velocity                .             .             .             .  .......            211 

Viscosity  of  Oil          .                          .  .     • '     .             .            .             .      203 

Volt          .             .                          ....  144-151-167-173 

Voltage  Most  Suitable          ,            .  .            .            .            .            .151 

Voltaic  Battery  Grouping           .          .  .  .            .            .             .             159 

"       Cell 174 

Couple    .  158 

Yyater  Tube  Boilers— Advantages             ,             .             .         '    .    ,  6-38-48 

Columns — Connections  '     .             .             .     •        .             .  42 

'       Velocity  of              .  ...            ..             .            .  .      ic,7 

Evaporated,  How        .            .            .            ;            ...  198 

Column  Pressure   .  .             .             .          -  .             ..  198-199 

'       Changed  to  Steam       .             .  .          .            , .           ...  198 

Watt — Value  of         .             .  .             .     •        .            .             .  145-151 

Weight — Units  of            .             .                          ...             .  214 

Wheatstone  Bridge                .  .             .'.....             .  .  .     172 

Wire — Capacity  of           .             .             .-            .             .            .             .  147-148 

Wires— Resistances  of          .  .         144-169-170-171-172-173-174-175-176 

Work — Definition            .......  199 

"        Foot  Lbs.  and  Velocity  .           '.            .            .            .  .      203 

^ero  Temperature  Absolute     .  .  "          .  201 


244 


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